Question 1

Syllogism – Disjoint classes and a subset Statements: 1) No paper is a pen. 2) No pen is a pencil. 3) All erasers are papers. Conclusions: I) Some papers are erasers. II) No pencil is an eraser. III) No pen is an eraser. IV) All papers are erasers.

Accepted Answer

Introduction / Context:We are given two universal negatives and one universal affirmative. The task is to infer which relationships must hold involving the subset 'erasers'.Given Data / Assumptions:Pa ∩ Pe = ∅ (no paper is a pen)Pe ∩ Pn = ∅ (no pen is a pencil)E ⊆ Pa (all erasers are papers)Concept / Approach:From E ⊆ Pa and Pa ∩ Pe = ∅, erasers cannot be pens. However, nothing here links pencils to papers or erasers, so be careful not to chain two disjointness statements in an invalid way (disjointness is not transitive).Step-by-Step Solution:III: No pen is an eraser. True, because any eraser is a paper, and no paper is a pen. Thus E ∩ Pe = ∅.II: No pencil is an eraser. Not forced. We have no premise connecting pencils to papers or erasers; 'no pen is a pencil' does not help.I: Some papers are erasers. This would require existence of at least one eraser. The premises do not guarantee that; some exams avoid asserting existence from 'all' statements. Hence I is not necessary.IV: All papers are erasers. Reverses the given subset; not supported.Verification / Alternative check:Construct a model with zero erasers. Then I is false, II is undetermined, III is trivially true, and IV is false. None of the option groupings matches this combination, so the meta-choice 'None of these' is correct.Why Other Options Are Wrong:A, B, C, D include II and/or I as necessary, which they are not, or claim all are true, which is incorrect.Common Pitfalls:Treating disjointness as transitive; presuming existence from a universal statement without support.Final Answer:None of these

Question 2

Syllogism – Combining two universal negatives with a particular fact Statements: 1) No man is the sky. 2) No sky is a road. 3) Some men are roads. Conclusions: I) No road is a man. II) No road is the sky. III) Some skies are men. IV) All roads are m…

Accepted Answer

Introduction / Context:This question tests careful reading of universal negatives alongside a 'some' statement. We must avoid drawing unwarranted converses and overgeneralizations.Given Data / Assumptions:M ∩ S = ∅ (no man is sky)S ∩ R = ∅ (no sky is road)∃ element in M ∩ R (some men are roads)Concept / Approach:Disjointness is symmetric, so from 'No sky is road' we can also say 'No road is sky' (II). But a particular overlap between men and roads refutes any universal separation of roads from men.Step-by-Step Solution:II: No road is sky. This follows immediately from 'No sky is road' (disjointness is mutual).I: No road is man. False because we are given some men are roads.III: Some skies are men. False because no man is sky.IV: All roads are men. Not forced; we know some roads are men, but not necessarily all.Verification / Alternative check:A quick model: pick one element that is both man and road. Keep sky disjoint from both men and roads. Then only II holds. Since there is no option that states 'Only II follows', the correct meta-answer is 'None of these' among the provided choices.Why Other Options Are Wrong:A: Claims none follow; II clearly follows.B, C, D: Each includes at least one false conclusion (I and/or III) or omits the exact needed combination.Common Pitfalls:Forgetting symmetry of disjointness; confusing 'some' with 'all'; assuming a universal without textual support.Final Answer:None of these

Question 3

Syllogism – Testing consequences of subset and disjointness Statements: 1) All buildings are windows. 2) No toy is a building. 3) Some tigers are toys. Conclusions: I) Some tigers are buildings. II) Some windows are tigers. III) All toys are tigers.…

Accepted Answer

Introduction / Context:Here we have one subset relationship and one universal negative, plus a particular fact about tigers and toys. The aim is to see whether any of the stated consequences must follow.Given Data / Assumptions:B ⊆ W (all buildings are windows)Toy ∩ B = ∅ (no toy is a building)∃ element in Tiger ∩ Toy (some tigers are toys)Concept / Approach:The particular fact ties tigers to toys, and the universal negative separates toys from buildings. The subset only says buildings lie within windows, not that all windows are buildings.Step-by-Step Solution:I: Some tigers are buildings. False, because those tigers picked from Tiger ∩ Toy are toys, and toys are not buildings.II: Some windows are tigers. Not guaranteed. We know nothing that links windows to toys or tigers; windows may exist outside buildings, but no necessary overlap with tigers is given.III: All toys are tigers. False; we only know some tigers are toys, not that all toys are tigers.IV: Some windows are toys. Not necessary because toys are disjoint from buildings, and buildings are the only guaranteed subset of windows; toys may or may not be windows via other routes, which are not specified.Verification / Alternative check:A model with no overlap between windows (outside buildings) and toys satisfies all premises and makes II and IV false, confirming that none of the conclusions must follow.Why Other Options Are Wrong:B, C, D, E each assert at least one conclusion that is not necessary under the premises.Common Pitfalls:Assuming all windows are buildings (converse error); assuming toys must connect to windows or buildings beyond what is stated.Final Answer:None follows

Question 4

Syllogism – Particular and universal statements together Statements: 1) Some papers are cats. 2) All cats are bats. 3) No bat is a horse. Conclusions: I) Some papers are horses. II) No horse is a cat. III) Some bats are papers. IV) All papers are ba…

Accepted Answer

Introduction / Context:This is a classic chain with one particular premise. By following the chain carefully, you can see which subsets must overlap and which cannot.Given Data / Assumptions:P ∩ C ≠ ∅ (some papers are cats)C ⊆ B (all cats are bats)B ∩ H = ∅ (no bat is a horse)Concept / Approach:Combine the particular statement with the subset inclusion: the particular papers that are cats must also be bats. With the universal negative regarding horses, you can infer impossibilities and necessary inclusions.Step-by-Step Solution:III: Some bats are papers. The papers that are cats are also bats, so at least one bat is a paper. Hence III follows.II: No horse is a cat. Since all cats are bats and no bat is a horse, cats and horses are disjoint. Therefore II follows.I: Some papers are horses. Not supported. The particular papers known to us (those which are cats) cannot be horses, and we do not know anything about other papers being horses.IV: All papers are bats. Overgeneralization. Only the papers that are cats must be bats; other papers need not be.Verification / Alternative check:Draw a Venn sketch: place a non-empty overlap of papers and cats within bats; keep horses disjoint from bats. Then II and III hold, I and IV do not.Why Other Options Are Wrong:A and D include I, which does not follow. C includes IV, which overstates the relationship. E claims all, which is incorrect.Common Pitfalls:Assuming that a property of some papers extends to all papers; ignoring that a universal negative about bats and horses blocks any cat–horse overlap.Final Answer:Only II and III follow

Question 5

Logical syllogism — test what necessarily follows from “some” statements  Statements: • Some tapes are discs. • Some discs are cassettes. • Some cassettes are songs. Conclusions to evaluate: I. Some songs are discs. II. Some cassettes are tapes. III…

Accepted Answer

Introduction / Context:This verbal reasoning problem checks your command of categorical logic with “some” statements. When premises use qualifiers like some, you must be careful not to over infer intersections that are not guaranteed. The task is to decide which, if any, of the proposed conclusions must be true based only on the given statements.Given Data / Assumptions:Some tapes are discs.Some discs are cassettes.Some cassettes are songs.No information is given about universal inclusion (all) or exclusions (no) among these four sets beyond the three some relations.Concept / Approach:In set logic, “some A are B” only asserts a nonempty overlap between A and B. Two separate some statements that share a middle term (for example tapes–discs and discs–cassettes) do not force an overlap between the outer classes (tapes and cassettes). Chains of some statements cannot be freely transitive without extra data. A conclusion follows only if it holds in every model consistent with the premises.Step-by-Step Solution:Test I (Some songs are discs): You can construct a model where the cassettes that are songs are disjoint from the discs that overlap tapes. Therefore I is not necessary.Test II (Some cassettes are tapes): From some tapes–discs and some discs–cassettes, the overlap could be with different parts of discs. Thus tapes and cassettes may still be disjoint. II does not follow.Test III (Some songs are tapes): Similarly, the some cassettes–songs overlap may be with cassettes that are not discs, whereas tapes overlap a different portion of discs. Hence III does not follow.Test IV (No song is a disc): A total exclusion is even stronger and is not implied; songs could be among the same cassettes that are also discs. IV does not follow.Verification / Alternative check:Draw Venn diagrams or pick concrete elements: let D1 overlap T (tapes) but not C (cassettes), let D2 overlap C but not T, and let S overlap a subset of C disjoint from D2. All premises remain true while I–IV can all be false. This proves none of the conclusions is forced.Why Other Options Are Wrong:Only either I or IV follows: neither I nor IV follows.Only either II or IV follows: neither follows.Only III and IV follow: both are unsupported.Only III and either II or IV follows: again, none of those are necessary.Common Pitfalls:Assuming transitivity for some; believing that three some links guarantee end-to-end overlap; treating absence of information as evidence of exclusion.Final Answer:None of these

Question 6

Syllogism with negative and universal claims — infer only what is necessary  Statements: • No table is fruit. • No fruit is window. • All windows are chairs. Conclusions to evaluate: I. No window is table. II. No chair is fruit. III. No chair is tab…

Accepted Answer

Introduction / Context:This question blends universal negative statements (no A is B) with a universal affirmative (all windows are chairs). The goal is to avoid illicit conversions and not read more than what is stated. We must determine which conclusions are logically compelled.Given Data / Assumptions:No table is fruit.No fruit is window.All windows are chairs.Nothing is said about non window chairs or any direct link between chairs and tables.Concept / Approach:From all windows are chairs we can infer chairs include all windows, but not that all chairs are windows (converse is invalid). From no fruit is window and all windows are chairs, we only know windows are not fruits. Whether chairs other than windows are fruits remains unknown. Similarly, no link is provided between windows and tables.Step-by-Step Solution:Conclusion I (No window is table): Premises do not connect windows with tables. It is consistent that some windows are tables unless forbidden elsewhere; no such bar exists. Hence I does not follow.Conclusion II (No chair is fruit): Only windows are guaranteed to be chairs, and windows are not fruits. Chairs beyond windows could still be fruits. II does not follow.Conclusion III (No chair is table): No rule connects chairs and tables; III does not follow.Conclusion IV (All chairs are windows): This is the illicit converse of all windows are chairs and is invalid. IV does not follow.Verification / Alternative check:Create a model: let W be a subset of C (chairs). Let some chairs outside W be fruits. Let some tables be non fruits. All premises hold, while I–IV all fail, proving none follows.Why Other Options Are Wrong:Only I and II / Only III and IV / All follow: each asserts conclusions not guaranteed by the premises.None of these: option A already states the correct outcome that none follows.Common Pitfalls:Converting “all windows are chairs” to “all chairs are windows”; assuming properties of the subset (windows) extend to the superset (chairs).Final Answer:None follows

Question 7

Layered category reasoning — track subset chains carefully  Statements: • All jungles are buses. • All books are buses. • All fruits are books. Conclusions to evaluate: I. Some fruits are jungles. II. Some buses are books. III. Some buses are jungle…

Accepted Answer

Introduction / Context:This is a classic chain of subset relations. The key skill is keeping track of inclusions and identifying which existential statements are warranted by the universals given the usual non emptiness convention in such exam problems.Given Data / Assumptions:All jungles are buses (J ⊆ B).All books are buses (Bk ⊆ B).All fruits are books (F ⊆ Bk).Standard test convention assumes non empty categories mentioned (there are fruits, books, jungles).Concept / Approach:Transitivity of subset works for universal affirmatives: if F ⊆ Bk and Bk ⊆ B, then F ⊆ B. From universal plus non emptiness, we may infer some statements like some buses are books (pick any book, it is a bus). However, without an explicit link between fruits and jungles, we cannot infer their overlap.Step-by-Step Solution:Conclusion I (Some fruits are jungles): There is no premise connecting fruits to jungles. They could be disjoint while all premises remain true. So I does not follow.Conclusion II (Some buses are books): Since all books are buses and books are assumed to exist (because fruits are books), at least one bus is a book. II follows.Conclusion III (Some buses are jungles): Since all jungles are buses and jungles are assumed to exist, at least one bus is a jungle. III follows.Conclusion IV (All fruits are buses): From F ⊆ Bk and Bk ⊆ B, transitivity gives F ⊆ B. IV follows.Verification / Alternative check:Set diagram: place F inside Bk inside B. Place J also inside B but not intersecting F to see that I need not hold, while II, III, IV remain valid.Why Other Options Are Wrong:Only I, II and III: includes I which is not necessary and misses IV which is certain.Only I, II and IV: again, wrongly includes I and omits III.All follow: wrongly accepts I.None of these: incorrect because II, III, IV do follow.Common Pitfalls:Assuming overlap between unrelated subclasses (fruits and jungles); forgetting to apply subset transitivity; overlooking that universal statements with existing classes permit some conclusions.Final Answer:Only II, III and IV follow

Question 8

Letter-ordering (forms a meaningful English word): Given letters with fixed positions — V(1) A(2) R(3) S(4) T(5) E(6) — which index sequence produces a valid word?

Accepted Answer

Introduction / Context:This item asks you to pick an index order that, when applied to the fixed letter slots V(1), A(2), R(3), S(4), T(5), E(6), yields a meaningful English word. Such questions test visual sequencing, anagram recognition, and careful mapping of indices to letters.Given Data / Assumptions:Fixed mapping: 1→V, 2→A, 3→R, 4→S, 5→T, 6→E.Each option lists a permutation of indices. We must read letters in that order and check if the result is an English word.Exactly one option should give a standard, dictionary word.Concept / Approach:Translate each proposed index list into letters and validate the result. A quick mental target is “STARVE,” a well-known word that can be formed from these letters.Step-by-Step Solution:Check option C: 4,5,2,3,1,6 → S(4), T(5), A(2), R(3), V(1), E(6) = “STARVE”.Other options produce non-words on inspection.Verification / Alternative check:Reverse-check: “STARVE” letters map back to indices S→4, T→5, A→2, R→3, V→1, E→6, matching option C exactly.Why Other Options Are Wrong:They yield strings that do not correspond to common English words (e.g., “ARV…” or “V…” starts that fail dictionary checks).Common Pitfalls:Mixing up the fixed letter-index mapping or reading indices as positions to rearrange (they are read order, not swap instructions).Final Answer:4, 5, 2, 3, 1, 6

Question 9

Letter-ordering (forms a meaningful English word): Given letters with fixed positions — A(1) L(2) I(3) R(4) E(5) M(6) C(7) — which index sequence produces a valid word?

Accepted Answer

Introduction / Context:We must pick an index order that turns A(1), L(2), I(3), R(4), E(5), M(6), C(7) into a standard English word.Given Data / Assumptions:Mapping: 1→A, 2→L, 3→I, 4→R, 5→E, 6→M, 7→C.Exactly one option should create a dictionary word.Concept / Approach:Try to recognize a target quickly. “MIRACLE” fits these letters.Step-by-Step Solution:Option A: 6,3,4,1,7,2,5 → M(6), I(3), R(4), A(1), C(7), L(2), E(5) = “MIRACLE”.Verification / Alternative check:Spelling check confirms “MIRACLE.” Other options scramble the order into non-words.Why Other Options Are Wrong:They fail to produce a common English word.Common Pitfalls:Swapping letter-index pairs or overlooking the simple pattern that spells “MIRACLE.”Final Answer:6, 3, 4, 1, 7, 2, 5

Question 10

Letter-ordering (forms a meaningful English word): Given letters with fixed positions — T(1) P(2) A(3) F(4) N(5) F(6) I(7) L(8) I(9) — which index sequence produces a valid word?

Accepted Answer

Introduction / Context:Pick the index order that converts the given sequence into a meaningful word.Given Data / Assumptions:Index map: 1→T, 2→P, 3→A, 4→F, 5→N, 6→F, 7→I, 8→L, 9→I.One option yields a standard English word.Concept / Approach:Look for high-frequency legal targets. The letters suggest “PLAINTIFF.”Step-by-Step Solution:Option D: 2,8,3,9,5,1,7,4,6 → P, L, A, I, N, T, I, F, F = “PLAINTIFF”.Verification / Alternative check:“PLAINTIFF” is correctly spelled and fits all letters exactly once.Why Other Options Are Wrong:They create non-words or malformed strings.Common Pitfalls:Missing the double-F ending or mixing I/L positions.Final Answer:2, 8, 3, 9, 5, 1, 7, 4, 6

Question 11

Letter-ordering (forms a meaningful English word): Given letters with fixed positions — E(1) M(2) I(3) H(4) T(5) R(6) — which index sequence produces a valid word?

Accepted Answer

Introduction / Context:Select the ordering that turns E(1), M(2), I(3), H(4), T(5), R(6) into a dictionary word.Given Data / Assumptions:Mapping: 1→E, 2→M, 3→I, 4→H, 5→T, 6→R.Concept / Approach:Common candidate here is “HERMIT.”Step-by-Step Solution:Option B: 4,1,6,2,3,5 → H, E, R, M, I, T = “HERMIT”.Verification / Alternative check:Spelling is correct; other options fail to form a standard word.Why Other Options Are Wrong:They create jumbles with no lexical meaning.Common Pitfalls:Overlooking the H→E start for “HER…”.Final Answer:4, 1, 6, 2, 3, 5

Question 12

Letter-ordering (forms a meaningful English word): Given letters with fixed positions — X(1) O(2) M(3) L(4) C(5) P(6) E(7) — which index sequence produces a valid word?

Accepted Answer

Introduction / Context:Pick an index sequence to form a known English word from the fixed sequence X(1), O(2), M(3), L(4), C(5), P(6), E(7).Given Data / Assumptions:Indices map 1→X, 2→O, 3→M, 4→L, 5→C, 6→P, 7→E.Concept / Approach:The natural target is “COMPLEX.”Step-by-Step Solution:Option D: 5,2,3,6,4,7,1 → C, O, M, P, L, E, X = “COMPLEX”.Verification / Alternative check:Compare letters and positions; all letters used once; spelling correct.Why Other Options Are Wrong:They yield non-words or misordered strings.Common Pitfalls:Swapping PX or ending on E instead of X.Final Answer:5, 2, 3, 6, 4, 7, 1

Question 13

Letter-ordering (forms a meaningful English word): Given letters with fixed positions — E(1) C(2) O(3) T(4) I(5) H(6) T(7) Y(8) H(9) O(10) N(11) P(12) A(13) — which index sequence produces a valid word?

Accepted Answer

Introduction / Context:We again select the index order producing a real English word from a longer letter sequence.Given Data / Assumptions:Map: 1→E, 2→C, 3→O, 4→T, 5→I, 6→H, 7→T, 8→Y, 9→H, 10→O, 11→N, 12→P, 13→A.Concept / Approach:These letters can spell “HYPOTHECATION,” a finance/legal term meaning pledge of property as security.Step-by-Step Solution:Option D yields indices 9,8,12,3,7,6,1,2,13,4,5,10,11 → H, Y, P, O, T, H, E, C, A, T, I, O, N = “HYPOTHECATION”.Verification / Alternative check:Spelling matches the standard word exactly; every letter is drawn from the provided positions in correct frequency.Why Other Options Are Wrong:They do not spell a standard dictionary word.Common Pitfalls:Misplacing the two H’s or the sequence “…THECATION”.Final Answer:9, 8, 12, 3, 7, 6, 1, 2, 13, 4, 5, 10, 11

Question 14

Letter-ordering (forms a meaningful English word): Given letters with fixed positions — R(1) T(2) A(3) N(4) U(5) E(6) — which index sequence produces a valid word?

Accepted Answer

Introduction / Context:Pick the ordering that gives a meaningful English word from R(1), T(2), A(3), N(4), U(5), E(6).Given Data / Assumptions:Index map: 1→R, 2→T, 3→A, 4→N, 5→U, 6→E.Concept / Approach:Target word is “NATURE.”Step-by-Step Solution:Option C: 4,3,2,5,1,6 → N, A, T, U, R, E = “NATURE”.Verification / Alternative check:Spelling and letter counts check out.Why Other Options Are Wrong:They fail to produce any standard word.Common Pitfalls:Swapping T and U or ending on R instead of E.Final Answer:4, 3, 2, 5, 1, 6

Question 15

Logic consistency: Statement: “When it rains, plants get greener.” From the following four atomic statements, select the pair that is logically consistent with the given rule. (i) It rained.  (ii) It did not rain.  (iii) Plants got greener.  (iv) Pl…

Accepted Answer

Introduction / Context:The rule “When it rains, plants get greener” is an implication: Rain → Greener. A consistent pair should not violate this implication.Given Data / Assumptions:Interpretation is material implication: if Rain is true, Greener must be true.Contrapositive (not used directly): If not Greener, then not Rain.Concept / Approach:Pick the pair that affirms Rain and Greener together (which certainly satisfies the rule).Step-by-Step Solution:(i) It rained; (iii) Plants got greener → matches Rain → Greener.Verification / Alternative check:Pairs with “It rained” and “plants did not get greener” would contradict the rule; pairs without Rain do not test the rule.Why Other Options Are Wrong:(i)(ii) is self-contradictory; (iii)(i) is the same facts in reverse order but option text expects the direct pair (we select the explicit cause-then-effect pair); (iv)(i) contradicts the contrapositive.Common Pitfalls:Assuming the converse (Greener → Rain) which is not stated.Final Answer:(i) (iii)

Question 16

Logic consistency: Statement: “When it is not raining, peacocks dance.” Select the pair that is consistent with this rule. (i) Peacocks are dancing.  (ii) It is raining.  (iii) Peacocks are not dancing.  (iv) It is not raining.

Accepted Answer

Introduction / Context:The rule is ¬Rain → Dance (if it is not raining, peacocks dance).Given Data / Assumptions:The rule does not assert anything about what happens when it is raining.Concept / Approach:The safest consistent pair is to assert the condition and its guaranteed outcome: not raining and dancing.Step-by-Step Solution:(iv) It is not raining; (i) Peacocks are dancing → consistent with ¬Rain → Dance.Verification / Alternative check:Pairs involving (ii) “It is raining” with dancing are not ruled out by the rule but do not follow from it; the test typically wants the direct cause→effect match.Why Other Options Are Wrong:They either state unrelated or potentially contradictory facts given the intended direction of implication.Common Pitfalls:Confusing “when” with biconditional (“if and only if”).Final Answer:(iv) (i)

Question 17

Logic consistency (exclusive alternative implied): Statement: “Madhu either runs or walks.” Select the pair that is most consistent with the intended meaning. (i) Madhu runs.  (ii) Madhu walks.  (iii) Madhu does not run.  (iv) Madhu does not walk.

Accepted Answer

Introduction / Context:“Either … or …” is typically read as an exclusive alternative in reasoning puzzles unless otherwise specified.Given Data / Assumptions:Exactly one of {runs, walks} is true.Concept / Approach:The combination “does not run” and “walks” satisfies exclusivity.Step-by-Step Solution:Pick (iii) (ii): not run, and walks.Verification / Alternative check:(i)(ii) asserts both true, contradicting exclusivity; (iii)(iv) asserts both false, also contradicting “either.”Why Other Options Are Wrong:They violate exclusive-or or create contradictions.Common Pitfalls:Reading “either … or …” as inclusive.Final Answer:(iii) (ii)

Question 18

Logic consistency: Statement: “She does not get a call when there is a strong network.” Select the pair that does not violate the rule and is jointly consistent. (i) She gets a call.  (ii) There is a strong network.  (iii) There is no network.  (iv)…

Accepted Answer

Introduction / Context:The statement is unusual but reads as: StrongNetwork → NoCall. We need a pair that is consistent with this rule (i.e., does not contradict it).Given Data / Assumptions:If strong network is present, a call does not occur.The converse (NoCall → StrongNetwork) is not implied.Concept / Approach:Pairs not mentioning strong network can still be consistent provided they do not force a contradiction.Step-by-Step Solution:(iii) No network; (iv) She does not get a call → consistent, as the rule talks only about the strong-network case.Verification / Alternative check:(ii)(i) would violate the rule (strong network but she gets a call). (iv)(ii) tries to infer the converse, which is not logically valid.Why Other Options Are Wrong:They either contradict the implication or rely on unstated converses.Common Pitfalls:Equating “no call” with “strong network” (converse error).Final Answer:(iii) (iv)

Question 19

Logic consistency: Statement: “When Baby wears only sari, Crazy goes bungee jumping.” Select the consistent pair. (i) Baby wore sari.  (ii) Crazy went bungee jumping.  (iii) Baby wore only sari.  (iv) Crazy did not go bungee jumping.

Accepted Answer

Introduction / Context:The rule is: OnlySari → Jumps (if Baby wore only sari, then Crazy surely jumped).Given Data / Assumptions:“Only sari” is stronger than “sari.”Concept / Approach:Pick the pair reflecting the condition and its effect (order may be effect–cause in options but should be consistent).Step-by-Step Solution:(ii) Crazy jumped; (iii) Baby wore only sari → consistent with the implication.Verification / Alternative check:Note that (i) “wore sari” alone is not sufficient to guarantee the jump (it must be “only sari”).Why Other Options Are Wrong:They either negate the guaranteed outcome or use an insufficient antecedent.Common Pitfalls:Confusing “sari” with “only sari.”Final Answer:(ii) (iii)

Question 20

Logic consistency: Statement: “If Raman works the late-night shift, his sister makes coffee.” Select the consistent pair. (i) Raman works the late-night shift.  (ii) Raman works the morning shift.  (iii) His sister makes coffee.  (iv) His sister mak…

Accepted Answer

Introduction / Context:The implication is LateNight → Coffee.Given Data / Assumptions:Morning shift info is irrelevant to the implication.Concept / Approach:Pick the antecedent and its consequent together.Step-by-Step Solution:(i) Late-night; (iii) Coffee → matches the rule.Verification / Alternative check:No contradictions arise.Why Other Options Are Wrong:They either negate or introduce unrelated facts.Common Pitfalls:Assuming “morning shift” tells us anything about coffee.Final Answer:(i) (iii)

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