Question 1

Syllogism — Evaluate conclusions from the statements: Statements: Some peons are poor. X is poor. Conclusions: I) X is a peon. II) X has a large family. Decide which conclusion(s), if any, logically follow(s) from the given statements.

Accepted Answer

Introduction / Context: In categorical syllogisms, we examine whether a conclusion must be true given the premises. The premises here are: (1) Some peons are poor, and (2) X is poor. We test whether the conclusions about X's being a peon or having a large family necessarily follow. Given Data / Assumptions: Premise 1: Some peons are poor (at least one individual is both peon and poor). Premise 2: X is poor (membership only in the set 'poor' is known for X). No information is provided about X's family size. Concept / Approach: Translate to sets: P = peons, R = poor. Premise 1 says P ∩ R is non-empty. Premise 2 says X ∈ R. A conclusion about X ∈ P would require that all poor are peons (R ⊆ P) or at least that this particular X is identified within P ∩ R, which is not given. Family-size claims are extraneous to premises. Step-by-Step Solution: Step 1: From 'Some peons are poor' infer existence of at least one member in P ∩ R.Step 2: From 'X is poor' infer only X ∈ R.Step 3: There is no rule binding all poor to be peons, so X ∈ P cannot be forced.Step 4: No premise mentions family size, so nothing can be concluded about X's family. Verification / Alternative check: If we construct a model where some peons are poor and there are other poor people who are not peons, X could be one of those non-peon poor. Thus I need not hold. Family size data is absent, so II cannot hold. Why Other Options Are Wrong: Only I follows: invalid; X need not be a peon. Only II follows: invalid; no family information. Both follow: invalid for reasons above. Common Pitfalls: Confusing 'some' with 'all' and assuming properties about X just because X shares a broad set label. Also, adding facts (like family size) not present in premises. Final Answer: Neither I nor II follows.

Question 2

Syllogism — Test conclusions from: (A) All basketball players are tall men. (B) All basketball players are athletes. Conclusions: I) All tall men are basketball players. II) All athletes are basketball players.

Accepted Answer

Introduction / Context: We are given two universal premises about 'basketball players': they are tall men and they are athletes. We must check whether conclusions about all tall men or all athletes being basketball players necessarily follow. Given Data / Assumptions: A: BasketballPlayers ⊆ TallMen. B: BasketballPlayers ⊆ Athletes. No premise relating Athletes and TallMen directly, nor reversing subset relations. Concept / Approach: Universal statements of the form 'All X are Y' establish X ⊆ Y. Their converses ('All Y are X') are not valid without extra premises. Thus from X ⊆ Y we cannot conclude Y ⊆ X. Step-by-Step Solution: Step 1: From A, BasketballPlayers ⊆ TallMen; does not imply TallMen ⊆ BasketballPlayers.Step 2: From B, BasketballPlayers ⊆ Athletes; does not imply Athletes ⊆ BasketballPlayers.Step 3: Therefore, neither I nor II is forced by the premises. Verification / Alternative check: Construct a model where many tall men are not players, and many athletes are not players; the premises still hold. Hence the conclusions fail in at least one model, so they do not logically follow. Why Other Options Are Wrong: I alone can be drawn: converse error. II alone can be drawn: converse error. Both can be drawn: both are converses and invalid. Common Pitfalls: Assuming 'All X are Y' implies 'All Y are X'. It does not unless explicitly stated. Final Answer: Both cannot be drawn.

Question 3

Syllogism — Given: Some doctors are teachers. All teachers are counsellors. Determine which conclusions follow: I) Some counsellors are not teachers. II) Some doctors are counsellors.

Accepted Answer

Introduction / Context: This problem involves set inclusions. We must check if the provided conclusions necessarily follow from the premises about doctors, teachers, and counsellors. Given Data / Assumptions: Premise 1: Some doctors are teachers (Doctors ∩ Teachers is non-empty). Premise 2: All teachers are counsellors (Teachers ⊆ Counsellors). Concept / Approach: Use transitivity of subset and existence from 'some'. If some doctors are teachers and all teachers are counsellors, then those doctor-teachers are counsellors, implying 'Some doctors are counsellors'. However, 'Some counsellors are not teachers' introduces a negative that is not supported by the premises. Step-by-Step Solution: Step 1: From Premise 1, choose an element d with d ∈ Doctors and d ∈ Teachers.Step 2: From Premise 2, Teachers ⊆ Counsellors, hence d ∈ Counsellors.Step 3: Therefore, some doctors are counsellors (existential conclusion true).Step 4: There is no information guaranteeing the existence of counsellors outside Teachers, so 'Some counsellors are not teachers' is not necessary. Verification / Alternative check: A model where Teachers = Counsellors makes I false but II true. As long as at least one doctor is a teacher, II follows. Why Other Options Are Wrong: Only I follows: unsupported; could be false if Counsellors equals Teachers. Both follow: I is not necessary. Neither follows: II must follow. Common Pitfalls: Assuming a larger set necessarily contains elements outside a subset without evidence. Final Answer: Only conclusion II follows.

Question 4

Syllogism — Evaluate: (a) All politicians are intelligent. (b) Some women are politicians. Conclusions: I) Some women are intelligent. II) All intelligent persons are women.

Accepted Answer

Introduction / Context: The task is to deduce what must be true from two premises linking politicians, intelligence, and women. Given Data / Assumptions: (a) Politicians ⊆ Intelligent. (b) Some Women are Politicians (Women ∩ Politicians non-empty). Concept / Approach: Chain subsets: if some women are politicians, and all politicians are intelligent, then those women are intelligent. Conversely asserting that all intelligent persons are women reverses the direction without support. Step-by-Step Solution: Step 1: Pick w with w ∈ Women ∩ Politicians.Step 2: From (a), Politicians ⊆ Intelligent, hence w ∈ Intelligent.Step 3: Therefore, 'Some women are intelligent' holds.Step 4: Nothing implies Intelligent ⊆ Women, so II fails. Verification / Alternative check: Countermodel for II: there can be intelligent men; II is thus not necessary. Why Other Options Are Wrong: Only II follows: converse error. Both follow: II is false. Neither follows: I must follow. Common Pitfalls: Confusing subset direction in universal statements. Final Answer: Only conclusion I follows.

Question 5

Syllogism — Premises: (A) Some cats are dogs. (B) Some dogs are stones. Conclusions: I) No cat is a stone. II) All dogs are stones. III) Some stones are cats. IV) No dog is a cat. Determine which, if any, follow.

Accepted Answer

Introduction / Context: We have two existential premises involving overlaps among cats, dogs, and stones. We must determine which stronger universal/negative or existential cross-set conclusions are forced. Given Data / Assumptions: Some Cats are Dogs (Cats ∩ Dogs ≠ ∅). Some Dogs are Stones (Dogs ∩ Stones ≠ ∅). No universal statements are given. Concept / Approach: From two separate 'some' statements, we cannot infer universal negatives or complete inclusion. Also, the overlapping dogs in the two premises need not be the same dogs. Step-by-Step Solution: I) 'No cat is a stone': This is too strong; it is compatible with the premises that some cat might be a stone or might not—nothing compels this universal negative.II) 'All dogs are stones': From 'some dogs are stones' we cannot generalize to all.III) 'Some stones are cats': This would require a specific triple overlap; premises allow but do not force it.IV) 'No dog is a cat': Contradicted by 'Some cats are dogs' which guarantees at least one dog-cat. Verification / Alternative check: Construct a model with three disjoint groups except for minimal overlaps exactly as stated: choose Dog1 ∈ Cats and Dog2 ∈ Stones with Dog1 ≠ Dog2. All premises true, but I, II, III, IV as evaluated above show none necessarily follow. Why Other Options Are Wrong: All listed combinations assert at least one of I–III–IV as true, which is not compelled. Common Pitfalls: Assuming the 'some' elements are the same individual across premises; overgeneralizing from existence to universals. Final Answer: None follows.

Question 6

Syllogism — Premises: A) All girls are proud. B) All proud will be humiliated one day. Inferences: I) All girls will be humiliated one day. II) Some girls will be humiliated one day.

Accepted Answer

Introduction / Context: Two chained universal premises are given. We assess whether the universal and existential versions of the resulting consequence follow. Given Data / Assumptions: A: Girls ⊆ Proud. B: Proud ⊆ HumiliatedOneDay. Concept / Approach: By transitivity, Girls ⊆ HumiliatedOneDay. From a universal inclusion, an existential statement 'Some girls ...' is also valid provided at least one girl exists; typical reasoning tests assume non-empty everyday categories. Step-by-Step Solution: Step 1: Combine A and B to get Girls ⊆ HumiliatedOneDay.Step 2: Conclusion I (universal) follows directly: all girls will be humiliated one day.Step 3: Conclusion II (existential) follows under the standard assumption that the class of girls is non-empty. Verification / Alternative check: Venn interpretation shows the Girls circle entirely inside Proud, which lies inside HumiliatedOneDay, making both I and II true in ordinary contexts. Why Other Options Are Wrong: Only I follows: ignores the routine existential inference. Only II follows: the universal also follows by subset transitivity. Neither: contradicted by the clear chain of subsets. Common Pitfalls: Forgetting that universal subset chains justify corresponding existential conclusions when the category is assumed non-empty. Final Answer: Both I & II follow.

Question 7

Syllogism — Premises: P) All men are women. Q) All women are crazy. Conclusions: I) All men are crazy. II) All the crazy are men. III) Some of the crazy are men. IV) Some of the crazy are women.

Accepted Answer

Introduction / Context: Two universal inclusions form a chain: Men ⊆ Women ⊆ Crazy. We test which conclusions necessarily follow in standard syllogistic settings common to reasoning exams. Given Data / Assumptions: P: Men ⊆ Women. Q: Women ⊆ Crazy. Routine assumption: the categories referenced (women, men) are not empty in ordinary-language reasoning problems. Concept / Approach: From chained subsets, we get Men ⊆ Crazy (I). The statement II claims the converse of Women ⊆ Crazy and is not supported. Statements III and IV are existential claims that follow if the relevant categories have at least one member. Step-by-Step Solution: I) Men ⊆ Crazy follows by transitivity: Men ⊆ Women and Women ⊆ Crazy.II) 'All crazy are men' is converse and does not follow; there can be crazy individuals who are not men.III) 'Some crazy are men' holds if any man exists since Men ⊆ Crazy.IV) 'Some crazy are women' holds if any woman exists since Women ⊆ Crazy. Verification / Alternative check: Construct any world with at least one woman and one man; then I, III, IV are true while II can be false if there exist crazy persons who are not men. Why Other Options Are Wrong: All follow: II fails. None follow: I certainly follows; III and IV follow under standard non-emptiness. Only II and III: II is wrong; IV also follows. Common Pitfalls: Confusing universal subset with its converse; overlooking typical non-emptiness assumptions used in exam syllogisms. Final Answer: Only I, III and IV follow.

Question 8

Syllogism — Premises: All Americans are English-speaking. No Eskimos are English-speaking. Conclusions: I) No Eskimos are Americans. II) No English-speakers are Eskimos.

Accepted Answer

Introduction / Context: This is a classic combination of a universal inclusion and a universal exclusion. We test whether each conclusion is forced. Given Data / Assumptions: Americans ⊆ EnglishSpeaking. Eskimos ∩ EnglishSpeaking = ∅ (equivalently, Eskimos ⊆ Not-EnglishSpeaking). Concept / Approach: If Eskimos share no members with English-speaking, and Americans are entirely within English-speaking, then Eskimos cannot be Americans. Also, 'No Eskimos are English-speaking' is equivalent to 'No English-speakers are Eskimos' in set-theoretic negation. Step-by-Step Solution: I) If Americans ⊆ EnglishSpeaking and Eskimos ∩ EnglishSpeaking = ∅, then Americans ∩ Eskimos = ∅. Hence no Eskimo is an American.II) The given exclusion is symmetric: A ∩ B = 0 implies B ∩ A = 0. So 'No English-speakers are Eskimos' follows directly. Verification / Alternative check: Venn diagrams show disjoint sets Eskimos and EnglishSpeaking, with Americans inside EnglishSpeaking; thus both conclusions hold. Why Other Options Are Wrong: Any option denying either I or II contradicts the disjointness and inclusion facts. Common Pitfalls: Forgetting that 'No A are B' is bidirectional in effect for membership claims (mutual exclusivity). Final Answer: Both I and II follow.

Question 9

Syllogism — Premises: (a) All frogs are tortoises. (b) No tortoise is a crocodile. Conclusions: I) No crocodile is a frog. II) No frog is a crocodile.

Accepted Answer

Introduction / Context: We combine a universal inclusion with a universal exclusion and test equivalent ways to state the same separation. Given Data / Assumptions: Frogs ⊆ Tortoises. Tortoises ∩ Crocodiles = ∅. Concept / Approach: If all frogs are tortoises and no tortoise is a crocodile, then frogs share the same exclusion from crocodiles. Statements I and II are equivalent rephrasings of that separation. Step-by-Step Solution: Step 1: Substitute Frogs within Tortoises: any frog is a tortoise.Step 2: From (b), that tortoise cannot be a crocodile.Step 3: Therefore no frog is a crocodile, and equivalently no crocodile is a frog. Verification / Alternative check: Venn diagram: Frogs inside Tortoises; Tortoises disjoint from Crocodiles; hence Frogs disjoint from Crocodiles. Why Other Options Are Wrong: Any option denying either I or II conflicts with the chain of inclusion and exclusion. Common Pitfalls: Missing that I and II are logically equivalent statements of disjointness. Final Answer: Both I and II follow.

Question 10

Syllogism — Premises: (a) All dogs are books. (b) All books are pictures. Conclusions: I) All dogs are pictures. II) All books are dogs. III) All pictures are dogs. IV) Some pictures are books.

Accepted Answer

Introduction / Context: Two nested universals are provided. We assess which of the four conclusions are compelled by these inclusions. Given Data / Assumptions: Dogs ⊆ Books. Books ⊆ Pictures. Ordinary-language assumption: the class 'Books' is non-empty, so existential statements about some members are acceptable in exam settings. Concept / Approach: Transitivity of subset gives Dogs ⊆ Pictures (I). The converses (II, III) are not supported. From Books ⊆ Pictures and non-emptiness, there exists at least one entity that is both picture and book, validating IV. Step-by-Step Solution: I) Dogs ⊆ Books and Books ⊆ Pictures imply Dogs ⊆ Pictures — true.II) 'All books are dogs': converse, not implied.III) 'All pictures are dogs': overly strong; not implied.IV) 'Some pictures are books': if any book exists, that book is a picture — validates an existential. Verification / Alternative check: Take one concrete book; it is also a picture, satisfying IV. Dogs being a subset of books ensures I. Why Other Options Are Wrong: Options that include II or III assume unsupported converses or overgeneralization. Common Pitfalls: Confusing subset direction and presuming equivalences where only one-way inclusion exists. Final Answer: Only I and IV follow.

Question 11

Syllogism — Premises: (a) All books are boxes. (b) All boxes are pens. (c) All pens are papers. Conclusions: I) Some papers are books. II) All books are papers. III) Some pens are books. IV) All boxes are books.

Accepted Answer

Introduction / Context: We have a chain of three universal inclusions. We must test a mix of universal and existential conclusions. Given Data / Assumptions: Books ⊆ Boxes. Boxes ⊆ Pens. Pens ⊆ Papers. Non-emptiness of Books for existential claims common in such tests. Concept / Approach: By transitivity, Books ⊆ Papers (II). Therefore, if any book exists, it is a paper, validating I. Also, Books ⊆ Pens, so some pens are books (III) under non-emptiness. IV (All boxes are books) reverses inclusion and is invalid. Step-by-Step Solution: Step 1: Compose subsets: Books ⊆ Boxes ⊆ Pens ⊆ Papers.Step 2: Conclude II: All books are papers — true.Step 3: Existentials: pick any book b; since b ∈ Papers, I is true. Also b ∈ Pens, so III is true.Step 4: IV asserts Boxes ⊆ Books, which is converse and not supported. Verification / Alternative check: A concrete chain example quickly confirms I, II, III while falsifying IV by allowing boxes that are not books. Why Other Options Are Wrong: Any option denying any of I–III or asserting IV conflicts with the chain. Common Pitfalls: Overlooking that existential conclusions require at least one member, typically assumed in exam problems. Final Answer: Only conclusions I, II and III follow.

Question 12

Syllogism — Premises: Some teachers are students. All students are girls. Conclusions: (I) All teachers are girls. (II) Some girls are teachers. (III) Some girls are students. (IV) All students are teachers.

Accepted Answer

Introduction / Context: We combine one existential and one universal premise and check which conclusions are necessary. Given Data / Assumptions: Some Teachers are Students. All Students are Girls. Concept / Approach: Let T = Teachers, S = Students, G = Girls. From 'Some T are S' and 'S ⊆ G', we infer (II) Some girls are teachers (those teacher-students are girls) and (III) Some girls are students (since some students exist). The universal claims (I) and (IV) are not supported. Step-by-Step Solution: Step 1: Pick x with x ∈ T ∩ S (from the 'some' statement).Step 2: From S ⊆ G, x ∈ G, so x ∈ G ∩ T — validates (II).Step 3: The existence of x ∈ S implies at least one student exists; since S ⊆ G, (III) holds.Step 4: There is no basis for 'All teachers are girls' (I) nor 'All students are teachers' (IV). Verification / Alternative check: Construct a model with many teachers not students and many girls not students; (II) and (III) remain forced while (I) and (IV) fail. Why Other Options Are Wrong: Options including (I) or (IV) assert unsupported universals. Common Pitfalls: Overgeneralizing from existence to universals; keep directions of subset relations straight. Final Answer: Only (II) and (III) follow.

Question 13

Syllogism — Premises: (a) All students like excursions. (b) Some students like laboratory experiments. Conclusions: (I) Students who like laboratory experiments also like excursions. (II) Some students do not like laboratory experiments but like exc…

Accepted Answer

Introduction / Context: We examine how a universal preference and a partial preference interact for conclusions about subsets of students. Given Data / Assumptions: All Students like Excursions. Some Students like LabExperiments. Concept / Approach: If every student likes excursions, then any particular subset of students (including those who like laboratory experiments) also like excursions. However, nothing in the premises asserts the existence of students who dislike laboratory experiments. Step-by-Step Solution: I) Let L be the set of students who like lab experiments. Since L ⊆ Students and all Students like excursions, it follows L ⊆ Excursions — conclusion I holds.II) The statement 'Some students do not like laboratory experiments but like excursions' introduces a negation not guaranteed by premises; the universal about excursions says nothing about lab-dislikers existing. Verification / Alternative check: A model where every student likes both excursions and lab experiments satisfies the premises but falsifies II, proving II does not necessarily follow. Why Other Options Are Wrong: Only II follows: contradicted by countermodel. Both follow: II is not necessary. Neither follows: I must follow by subset reasoning. Common Pitfalls: Assuming existence of complements (non-LabExperiment-likers) without evidence. Final Answer: Only Conclusion I follows.

Question 14

Syllogism — Premises: All hens are cocks. No cock is black. Conclusions: I) All cocks are hens. II) No hen is black.

Accepted Answer

Introduction / Context: This mixes a universal inclusion with a universal exclusion. We test each conclusion cautiously for direction and scope. Given Data / Assumptions: Hens ⊆ Cocks. Cocks ∩ Black = ∅. Concept / Approach: From Hens ⊆ Cocks and no Cock is Black, we infer Hens ∩ Black = ∅, i.e., no hen is black (II). The first conclusion reverses inclusion and is not supported. Step-by-Step Solution: Step 1: Substitute: any hen is a cock.Step 2: Since no cock is black, that hen cannot be black.Step 3: Hence II holds. I fails because 'All cocks are hens' is the converse and need not be true. Verification / Alternative check: Model where some cocks are not hens satisfies premises but falsifies I. Why Other Options Are Wrong: Any option including I treats converse as valid; it is not. Common Pitfalls: Confusing subset with set equality. Final Answer: Only conclusion II is valid.

Question 15

Syllogism — Analogy trap: Dogs have four legs. Tables have four legs. Conclusions: I) Tables are dogs. II) Dogs are tables. Determine which, if any, follow logically.

Accepted Answer

Introduction / Context: This question tests understanding of the fallacy of affirming the consequent / property-sharing. Two categories share a common attribute (four legs), but that alone does not license identity or inclusion conclusions. Given Data / Assumptions: Dogs have four legs. Tables have four legs. No premise states any inclusion or equivalence between Dogs and Tables. Concept / Approach: Having a common property does not imply class identity. Sets can overlap, be disjoint, or relate in many ways despite sharing attributes. Syllogistic conclusions require explicit subset/equality statements. Step-by-Step Solution: Step 1: Identify structure: From A has property L and B has property L, one cannot infer A = B nor A ⊆ B nor B ⊆ A.Step 2: Since no categorical link exists between Dogs and Tables, neither conclusion is forced. Verification / Alternative check: Counterexample: there exist tables without any relation to dogs besides leg count; likewise dogs are living beings, not furniture. Premises are satisfied; conclusions fail. Why Other Options Are Wrong: Options A, B: assert unjustified identity/inclusion. Option C: suggests one of them must hold, which is false. Common Pitfalls: Confusing shared attributes with class equivalence; ignoring the need for subset/equality premises. Final Answer: Neither I nor II follows.

Question 16

Syllogism — Premises: (a) All English movies are violent. (b) Some people like watching English movies. Conclusions: I) All people watching English movies like violence. II) All people who like violence watch English movies.

Accepted Answer

Introduction / Context: The premises link the content of English movies (violent) with a viewer subset (some people like watching English movies). The conclusions, however, make claims about viewers' preferences for violence or the source of all violence-likers. Given Data / Assumptions: All EnglishMovies are ViolentContent. Some People like watching EnglishMovies. No premise about people liking violence as a result of viewing. Concept / Approach: From 'All English movies are violent' we cannot infer anything about whether viewers like violence; they might watch for other reasons or dislike the violence yet still watch. Nor can we infer that all violence-likers watch English movies. Step-by-Step Solution: I) The premise is about movie content, not viewer preference; 'All viewers like violence' is unsupported.II) 'All violence-likers watch English movies' is an arbitrary universal connecting a preference group to a content class without evidence. Verification / Alternative check: Countermodels exist: viewers might enjoy plots despite violence; many who like violent content might prefer non-English media. Premises hold while both conclusions fail. Why Other Options Are Wrong: Any option asserting I or II introduces unsupported generalizations about audiences. Common Pitfalls: Confusing properties of content with viewers' attitudes; assuming causation or necessary preference from descriptive statements. Final Answer: Neither I nor II follows.

Question 17

Syllogism — Premises: (a) All cupboards are watches. (b) All watches are costly. Inferences: I) All cupboards are costly. II) Some costly things are cupboards.

Accepted Answer

Introduction / Context: Two universals create a subset chain. We must judge a universal and an existential inference that naturally arise from that chain. Given Data / Assumptions: Cupboards ⊆ Watches. Watches ⊆ Costly. Ordinary non-emptiness assumption for Cupboards to validate the existential. Concept / Approach: Transitivity yields Cupboards ⊆ Costly, establishing I. If at least one cupboard exists, it is costly, establishing II: 'Some costly things are cupboards'. Step-by-Step Solution: Step 1: Cupboards ⊆ Watches and Watches ⊆ Costly imply Cupboards ⊆ Costly.Step 2: Pick any cupboard (non-empty common-sense assumption); it is costly, proving the existential. Verification / Alternative check: Venn diagram placement confirms I and supports II with non-emptiness. Why Other Options Are Wrong: Options asserting only one or none ignore the straightforward chain. Common Pitfalls: Overlooking that an existential conclusion typically follows in exam settings when categories are assumed to exist. Final Answer: Both of them follow.

Question 18

Syllogism — Premises: All fruits are sweet. All sweet things are not good for health (i.e., no sweet thing is good for health). Conclusions: (I) All fruits are not good for health. (II) Some fruits are not good for health.

Accepted Answer

Introduction / Context: The phrase 'All sweet things are not good for health' is read in reasoning tests as 'No sweet thing is good for health' (a universal negative). Combined with 'All fruits are sweet', we test the two conclusions. Given Data / Assumptions: Fruits ⊆ Sweet. Sweet ⊆ NotGoodForHealth (equivalently, Sweet ∩ GoodForHealth = ∅). Fruits is a non-empty everyday category. Concept / Approach: By transitivity, Fruits ⊆ NotGoodForHealth, validating (I). If any fruit exists, then at least one fruit is not good for health, validating (II) as an existential consequence. Step-by-Step Solution: Step 1: Combine subsets: Fruits ⊆ Sweet and Sweet ⊆ NotGoodForHealth.Step 2: Conclude (I): All fruits are not good for health.Step 3: Non-emptiness yields (II): Some fruits are not good for health. Verification / Alternative check: Venn diagram confirms Fruits entirely lies within NotGoodForHealth, ensuring both universal and existential claims. Why Other Options Are Wrong: Any option denying either (I) or (II) conflicts with the derived inclusion and routine existential assumption. Common Pitfalls: Misreading 'all ... are not' as 'not all ... are', which is different in logic. Final Answer: Both (I) and (II) follow.

Question 19

Syllogism — Premises: (a) All pens are papers. (b) No eraser is a paper. Conclusions: I) No eraser is a pen. II) Some papers are pens.

Accepted Answer

Introduction / Context: We combine a universal inclusion with a universal exclusion and test both a universal negative and an existential conclusion. Given Data / Assumptions: Pens ⊆ Papers. Erasers ∩ Papers = ∅. Assume at least one pen exists (ordinary context) for the existential statement. Concept / Approach: If Erasers are disjoint from Papers and Pens are within Papers, then Erasers and Pens are disjoint (I). Also, if any pen exists, it is a paper, so at least some papers are pens (II). Step-by-Step Solution: I) Since Pens ⊆ Papers and Erasers ∩ Papers = ∅, it follows Erasers ∩ Pens = ∅ — no eraser is a pen.II) Pick any pen; it belongs to Papers, so 'Some papers are pens' holds. Verification / Alternative check: Venn diagrams depict Pens inside Papers and Erasers disjoint from Papers; both conclusions are immediate. Why Other Options Are Wrong: Options denying either I or II contradict the straightforward set relations. Common Pitfalls: Overlooking that an existential like 'some papers are pens' is justified by non-emptiness of Pens in typical exam logic. Final Answer: Both Conclusions I and II follow.

Question 20

Syllogism — Premises: (a) No girl is a parrot. (b) Sowmya is a girl. Conclusions: I) Sowmya is not a parrot. II) All girls are not Sowmya. Choose the conclusion(s) that follow.

Accepted Answer

Introduction / Context: This problem mixes a universal negative with an individual membership statement. We check which conclusions necessarily follow about the individual and about the class of girls. Given Data / Assumptions: No Girl is a Parrot (Girls ∩ Parrots = ∅). Sowmya ∈ Girls. Concept / Approach: Substituting the individual into the universal exclusion gives a certain negative about Sowmya. The second conclusion is a poorly formed generalization about identity of all girls with Sowmya and is not entailed by the premises. Step-by-Step Solution: I) Since Sowmya is a girl and no girl is a parrot, it follows Sowmya is not a parrot — valid.II) 'All girls are not Sowmya' is either trivially true in ordinary usage or ill-posed in formal syllogism, but it does not follow as a necessary logical consequence from the given premises; syllogism conclusions should relate directly to provided predicates. Verification / Alternative check: The only relation established is between Girls and Parrots, plus Sowmya's membership in Girls; nothing licenses a universal identity/inequality statement about all girls and Sowmya. Why Other Options Are Wrong: Any option granting II introduces a statement not derivable from the premises within standard syllogistic form. Common Pitfalls: Accepting statements that are colloquially obvious but not logically entailed by the given categorical premises. Final Answer: Only conclusion I follows.

Syllogism Questions