Question 1

Syllogism — University professors and doctorate degrees:  Statements: • All university professors have a doctorate degree. • X is a lady professor.  Conclusions to test: I. X does not have a doctorate degree. II. Only men professors have a doctorate…

Accepted Answer

Introduction / Context:This item tests categorical syllogism logic using a universal affirmative premise and an individual case. We must determine which stated conclusions necessarily follow from the given premises without adding outside facts or stereotypes.Given Data / Assumptions:All university professors have a doctorate degree (Professor ⊆ Doctorate-holders).X is a professor; additionally, X is a woman ('lady professor').Conclusions must be necessary, not merely plausible.Concept / Approach:Use set inclusion. From “All professors have a doctorate,” Professor ⊆ Doctorate. Instantiating for X yields X ∈ Doctorate. The premise says nothing about gender restrictions on doctorate holders.Step-by-Step Solution:1) From the universal rule, every professor must have a doctorate.2) Since X is a professor, X must have a doctorate.3) Conclusion I (“X does not have a doctorate”) contradicts step 2 and therefore does not follow.4) Conclusion II (“Only men professors have a doctorate”) introduces a gender restriction not present in the premises; it does not follow.Verification / Alternative check:Construct a model where professors of any gender all have doctorates; I becomes false and II is not forced.Why Other Options Are Wrong:“Only I follows” is wrong because I contradicts the premise. “Only II follows” is wrong because gender is never constrained. “Both follow” is impossible (I is false).Common Pitfalls:Assuming extra information about gender from social context, or misreading “All” as “Only men.”Final Answer:Neither I nor II follows.

Question 2

Syllogism — Philosophers and gender; Socrates case:  Statements: • All philosophers are men. • Socrates was a philosopher.  Conclusions to test: I. Socrates was a man. II. Women cannot become philosophers.  Choose which conclusion(s) necessarily fol…

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Introduction / Context:This syllogism examines what necessarily follows from a universal statement and a particular instance, avoiding any modal (“can/cannot become”) extrapolations unless they are strictly entailed.Given Data / Assumptions:Premise 1: All philosophers are men (Philosopher ⊆ Men).Premise 2: Socrates was a philosopher.We must test the two conclusions for necessity, not plausibility.Concept / Approach:Universal inclusion plus instance substitution. From “All P are M” and “Socrates is P,” we infer “Socrates is M.” The second conclusion introduces a modal impossibility claim about women, which is not warranted by the set inclusion alone.Step-by-Step Solution:1) From Premise 1, every philosopher is a man.2) From Premise 2, Socrates was a philosopher; substituting into step 1 yields Socrates was a man.3) Conclusion I is therefore necessary.4) Conclusion II (“Women cannot become philosophers”) goes beyond the given set inclusion; the premises do not speak about possibilities for women.Verification / Alternative check:Models can satisfy the premises while allowing or disallowing female philosophers without contradiction; hence II is not forced.Why Other Options Are Wrong:“Only II,” “Both,” and “Neither” each conflict with the necessary inference in I.Common Pitfalls:Reading social or historical context into logical form.Final Answer:Only I is valid.

Question 3

Syllogism — Monkeys, singing, and talking:  Statement: • No monkey can sing, but some monkeys can talk.  Which option states the correct logical consequence of the statement?

Accepted Answer

Introduction / Context:The problem checks understanding of universal negative statements (“No A are B”) and particular affirmative statements (“Some A are C”). We must choose the option that must be true, not one that merely sounds plausible.Given Data / Assumptions:No monkey can sing (Monkey ∩ Singer = ∅).Some monkeys can talk (∃ Monkey that are Talkers).Read “All monkeys can’t sing” as “No monkeys can sing.”Concept / Approach:Translate the English into set relations: the first clause is a universal exclusion, the second asserts existence of talkers among monkeys. Evaluate each option against these relations.Step-by-Step Solution:1) From “No monkey can sing,” it necessarily follows that no member of the Monkey set is in the Singer set.2) From “Some monkeys can talk,” at least one monkey is a Talker, but nothing is implied about all monkeys talking.3) (a) “Some monkeys may sing” contradicts the universal exclusion, so it is false.4) (b) “All monkeys can’t sing” restates the universal exclusion and must be true.5) (c) and (d) overgeneralize beyond the “some can talk” clause.Verification / Alternative check:A model with 10 monkeys, none singing, two talking, satisfies the statement and validates only option (b).Why Other Options Are Wrong:(a) contradicts the premise; (c) and (d) are not entailed.Common Pitfalls:Misreading “some” as “all,” or interpreting “can’t” ambiguously.Final Answer:All monkeys can't sing.

Question 4

Syllogism — Cows, chairs, and tables:  Statements: (A) No cow is a chair. (B) All chairs are tables.  Conclusions to test: I. Some tables are chairs. II. Some tables are cows. III. Some chairs are cows. IV. No table is a cow.  Select the correct opt…

Accepted Answer

Introduction / Context:This classic syllogism mixes a universal negative (No A are B) with a universal affirmative subset relation (All B are C). We must determine which conclusions are logically compelled without assuming information not given.Given Data / Assumptions:(A) No cow is a chair (Cow ∩ Chair = ∅).(B) All chairs are tables (Chair ⊆ Table).Concept / Approach:Map sets: Cow (Cw), Chair (Ch), Table (Tb) with Ch ⊆ Tb and Cw ∩ Ch = ∅. Evaluate each conclusion.Step-by-Step Solution:1) From (B), if chairs exist, those chairs are tables; hence “Some tables are chairs” is accepted in test convention (Conclusion I follows).2) II “Some tables are cows” has no support; nothing links tables (in general) to cows.3) III “Some chairs are cows” contradicts (A) and is false.4) IV “No table is a cow” overreaches: (A) only forbids cows from being chairs, not from being other kinds of tables.Verification / Alternative check:Let chairs exist and be a subset of tables; let cows be disjoint from chairs; allow tables that are not chairs. Only I is forced.Why Other Options Are Wrong:(a) and (b) include false parts; (d) is impossible; (e) fails because I does follow.Common Pitfalls:Converting “All B are C” to “All C are B,” or projecting disjointness beyond what is stated.Final Answer:Only I follows.

Question 5

Syllogism — Men, women, and intelligence:  Statements: • No man is intelligent. • All women are intelligent.  Conclusions to test: I. No man is a woman. II. No woman is a man.

Accepted Answer

Introduction / Context:This problem checks whether learners can combine a universal negative with a universal affirmative to infer disjointness between two classes.Given Data / Assumptions:No man is intelligent.All women are intelligent.Concept / Approach:If all women are intelligent, and no man is intelligent, then a person cannot simultaneously be a man and a woman; otherwise that person would be both intelligent (from being a woman) and not intelligent (from being a man), which is impossible. Hence the sets Man and Woman are disjoint.Step-by-Step Solution:1) From “All women are intelligent,” Woman ⊆ Intelligent.2) From “No man is intelligent,” Man ⊆ (not Intelligent).3) If any person were both Man and Woman, a contradiction would arise. Therefore, Man ∩ Woman = ∅.4) Conclusion I “No man is a woman” follows.5) Conclusion II “No woman is a man” is the same disjointness read in the other direction and also follows.Verification / Alternative check:Venn diagram: Woman inside Intelligent; Man entirely outside Intelligent; there is no overlap between Man and Woman.Why Other Options Are Wrong:Options stating only one or neither ignore that both directions of disjointness are equivalent.Common Pitfalls:Confusing “No man is intelligent” with “No intelligent person is a man” (illicit converse). The conclusion here is about gender sets, not the converse of premises.Final Answer:Both I & II follow.

Question 6

Syllogism — Carts, cars, and trains (transitivity):  Statements: • All carts are cars. • All cars are trains.  Conclusions to test: I. All carts are trains. II. All trains are carts.

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Introduction / Context:This item checks transitivity of set inclusion and the common mistake of reversing subset relations.Given Data / Assumptions:All carts are cars (Cart ⊆ Car).All cars are trains (Car ⊆ Train).Concept / Approach:Subset relations compose: if A ⊆ B and B ⊆ C, then A ⊆ C. But the converse (C ⊆ A) does not follow.Step-by-Step Solution:1) Chain the inclusions: Cart ⊆ Car and Car ⊆ Train implies Cart ⊆ Train.2) Conclusion I is therefore correct.3) Conclusion II “All trains are carts” would require Train ⊆ Cart, which is not warranted.Verification / Alternative check:Create a model: Carts (20) ⊆ Cars (100) ⊆ Trains (1000). Inclusion works one way only.Why Other Options Are Wrong:“Only II,” “Both,” and “Neither” contradict the valid transitive inclusion.Common Pitfalls:Converse error: assuming if A ⊆ B then B ⊆ A.Final Answer:Only conclusion I follows.

Question 7

Syllogism — Soldiers, citizens, and men:  Statements: • All soldiers are men. • Some citizens are soldiers.  Conclusions to test: I. Some citizens are men. II. All soldiers are citizens.

Accepted Answer

Introduction / Context:We combine a universal inclusion with a particular statement to test if a specific existential conclusion follows.Given Data / Assumptions:All soldiers are men (Soldier ⊆ Man).Some citizens are soldiers (∃ Citizen ∩ Soldier).Concept / Approach:From Soldier ⊆ Man, any soldier is a man. Since some citizens are soldiers, those very citizens are men; hence “Some citizens are men.” There is no basis to claim every soldier is a citizen.Step-by-Step Solution:1) Take the citizens who are soldiers (exist by premise).2) Map them through Soldier ⊆ Man to conclude they are men.3) Hence I follows. II claims Soldier ⊆ Citizen, which is not given.Verification / Alternative check:A model with 10 citizens who are soldiers among many other soldiers suffices; II fails.Why Other Options Are Wrong:(b) denies a necessary inference; (c) asserts an unsupported universal; (d) overstates.Common Pitfalls:Concluding a universal from a particular (“some” → “all”).Final Answer:Only I follows.

Question 8

Syllogism — Shoes, White, and Blue:  Statements: (a) Some shoes are White. (b) All White are Blue.  Conclusions to test: I. All shoes are Blue. II. Some shoes are Blue.

Accepted Answer

Introduction / Context:We must propagate a particular statement through a universal inclusion and see what minimum conclusion is guaranteed.Given Data / Assumptions:Some shoes are White (∃ Shoe ∩ White).All White are Blue (White ⊆ Blue).Concept / Approach:A subset mapping carries the “some” individuals along: the shoes that are White are necessarily Blue. But this does not make all shoes Blue.Step-by-Step Solution:1) Take the shoes that are White (exist by premise).2) Since White ⊆ Blue, those shoes are Blue.3) Therefore, “Some shoes are Blue” follows, while “All shoes are Blue” does not.Verification / Alternative check:Construct a model with many non-White shoes; I is false but II remains true.Why Other Options Are Wrong:(a) overgeneralizes; (c) and (d) misjudge necessity.Common Pitfalls:Confusing “some” with “all.”Final Answer:Only II follows.

Question 9

Syllogism — Files, numbered, and coded (ambiguity check):  Statements: • Some files are numbered. • All numbers are coded.  Which inference necessarily follows?

Accepted Answer

Introduction / Context:This question is intentionally tricky because the middle term shifts in sense. The first premise talks about “numbered (files)”; the second talks about “numbers” (as entities). Without a valid middle-term link, no conclusion about files and coding is forced.Given Data / Assumptions:Premise 1: Some files are numbered (∃ File that possess the property “numbered”).Premise 2: All numbers are coded (Number ⊆ Coded).No explicit link between “numbered (as a property of files)” and the noun class “numbers.”Concept / Approach:In syllogisms, the middle term must be used in the same sense. Here, “numbered” (adjective on files) is not the same as the noun class “numbers.” Thus, we cannot lawfully chain the premises.Step-by-Step Evaluation of Options:(a) “All files are coded” is an illicit generalization.(b) “All codes are filed” reverses and alters classes.(c) “Some files are not numbered” might be true, but it is not necessary from the premises; the first premise allows the rest of files to be numbered or not.(d) “No numbers have codes” contradicts Premise 2.Verification / Alternative check:A model where some files are numbered, numbers are coded, but no statement connects numbered-files to “coded” validates that none of (a)–(d) must hold.Final Answer:None of these.

Question 10

Syllogism — Goats, tigers, and lions (subset chains):  Statements: (a) All goats are tigers. (b) All tigers are lions.  Conclusions: I. All tigers are goats. II. All lions are tigers. III. No goat is a lion. IV. No lion is a goat.

Accepted Answer

Introduction / Context:Two universal inclusions are given. We must avoid both reversing inclusions and drawing unwarranted negatives.Given Data / Assumptions:Goat ⊆ Tiger.Tiger ⊆ Lion.Concept / Approach:From the chain, Goat ⊆ Lion. However, none of the listed conclusions matches this lawful result: (I) reverses inclusion; (II) wrongly claims Lion ⊇ Tiger in the reverse direction; (III) and (IV) assert disjointness that is contradicted or not implied.Step-by-Step Analysis:1) I “All tigers are goats” is false (wrong direction).2) II “All lions are tigers” is false (reverse inclusion).3) III “No goat is a lion” contradicts Goat ⊆ Lion.4) IV “No lion is a goat” is not implied; goats may be a proper subset of lions.Verification / Alternative check:Model: Goats (10) ⊆ Tigers (100) ⊆ Lions (1000). None of I–IV is forced true.Final Answer:None of the conclusions follow.

Question 11

Syllogism — Clerks, typists, and stenos (undetermined overlap):  Statements: (a) All clerks are typists. (b) Some typists are stenos.  Conclusions: I. Some stenos are clerks. II. No steno is a clerk. III. All typists are clerks. IV. All clerks are s…

Accepted Answer

Introduction / Context:We have one universal and one particular statement. The relationship between Clerks and Stenos is not fixed by the premises and may vary across models.Given Data / Assumptions:Clerk ⊆ Typist.Some Typists are Stenos (∃ Typist ∩ Steno).Concept / Approach:The intersection between Clerks and Stenos is undetermined: it could be empty or non-empty, consistent with the premises. Therefore, either “Some stenos are clerks” or “No steno is a clerk” will hold depending on the model, but neither is individually necessary.Step-by-Step Evaluation:1) If the “some typists that are stenos” are chosen from among clerks, then I is true.2) If they are chosen from non-clerk typists, then II is true.3) III “All typists are clerks” is unsupported (reverse inclusion).4) IV “All clerks are stenos” is also unsupported.Verification / Alternative check:Construct two models as above; one satisfies I, the other satisfies II; both keep the premises true.Final Answer:Either I or II follows.

Question 12

Syllogism — Players, educated, and cultured:  Statements: (a) All players are educated. (b) All educated are cultured.  Inferences: I. All players are cultured. II. All cultured persons are players.

Accepted Answer

Introduction / Context:This is a direct application of transitivity in set inclusion.Given Data / Assumptions:Player ⊆ Educated.Educated ⊆ Cultured.Concept / Approach:Chain the inclusions to get Player ⊆ Cultured. The converse Cultured ⊆ Player is not implied.Step-by-Step:1) Player ⊆ Educated and Educated ⊆ Cultured.2) Hence Player ⊆ Cultured (I follows).3) II incorrectly reverses the subset relation.Verification:Model with many cultured non-players refutes II while preserving the premises.Final Answer:Only inference I follows.

Question 13

Syllogism — Books, magazines, and novels:  Statements: • Some books are magazines. • Some magazines are novels.  Conclusions: (I) Some books are novels. (II) Some novels are magazines.

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Introduction / Context:This problem tests the difference between an overlap that is directly stated and a transitive overlap that is not guaranteed.Given Data / Assumptions:Some Books are Magazines.Some Magazines are Novels.Concept / Approach:From the second premise, the intersection Magazines ∩ Novels is non-empty; therefore (II) is immediately true. However, the first premise links Books to Magazines, not directly to Novels; the overlapping subsets could be disjoint inside Magazines, so (I) is not necessary.Step-by-Step:1) Use premise 2 to assert ∃ (Novels ∩ Magazines).2) Premise 1 asserts ∃ (Books ∩ Magazines).3) The two “some” groups inside Magazines need not be the same individuals, so (I) is not compelled.Verification:Draw Magazines as a large set with two disjoint pockets: one overlapping Books, another overlapping Novels. Premises hold while (I) fails.Final Answer:Only (II) follows.

Question 14

Syllogism — Animals, birds, and human beings:  Statements: • All animals are birds. • Some animals are human beings.  Conclusions: I. Some human beings are birds. II. Some birds are human beings.

Accepted Answer

Introduction / Context:We propagate a particular subset of Animals (those who are humans) through the universal inclusion Animals ⊆ Birds.Given Data / Assumptions:Animal ⊆ Bird.Some Animals are Human (∃ Animal ∩ Human).Concept / Approach:If all Animals are Birds, then the specific Animals that are Humans are necessarily Birds as well. Hence there exists at least one creature that is both Human and Bird. Both statements (I) and (II) express the same fact from different directions.Step-by-Step:1) From Animal ⊆ Bird, map members of Animal ∩ Human into Bird.2) Therefore, ∃ Human ∩ Bird (I).3) Symmetrically, ∃ Bird ∩ Human (II) holds by the same individuals.Verification:A Venn diagram confirms both conclusions are equivalent existential claims.Final Answer:Both I and II follow.

Question 15

Syllogism — Women, cats, and rats (subset chaining):  Statements: • All women are cats. • All cats are rats.  Conclusions: I. All women are rats. II. All rats are women.

Accepted Answer

Introduction / Context:Another transitivity check with attention to the direction of inclusion.Given Data / Assumptions:Woman ⊆ Cat.Cat ⊆ Rat.Concept / Approach:Chain inclusions to get Woman ⊆ Rat (I true). II reverses inclusion and is not warranted.Step-by-Step:1) Woman ⊆ Cat and Cat ⊆ Rat.2) Hence Woman ⊆ Rat (I follows).3) “All rats are women” would require Rat ⊆ Woman, which is not implied.Verification:Consider many rats that are not cats; premises do not forbid them.Final Answer:Only conclusion I follows.

Question 16

Syllogism — Books, trees, and lions (existential note):  Statements: (1) All books are trees. (2) All trees are lions.  Conclusions: I. All books are lions. II. All lions are books. III. All trees are books. IV. Some lions are books.

Accepted Answer

Introduction / Context:We examine inclusion chaining and the careful use of existential import for particular conclusions.Given Data / Assumptions:Book ⊆ Tree.Tree ⊆ Lion.Concept / Approach:From chaining, Book ⊆ Lion, so I holds. II and III reverse inclusions and are not compelled. IV “Some lions are books” follows under the standard test assumption that the subject classes are non-empty (i.e., books exist). This is commonly assumed in competitive-exam syllogisms when particulars are drawn from universals.Step-by-Step:1) Compose inclusions to derive Book ⊆ Lion.2) I follows immediately.3) II and III are unsupported reversals.4) If at least one Book exists (standard assumption), that entity is a Lion; hence IV holds.Verification:Let one book exist; then at least one lion is a book (the same entity), validating IV.Final Answer:Only I and IV follow.

Question 17

Syllogism — Clerks → Superintendents → Managers → Supervisors (long chain):  Statements: A. All clerks are superintendents. B. All superintendents are managers. C. All managers are supervisors.  Conclusions: I. All supervisors are clerks. II. Some c…

Accepted Answer

Introduction / Context:A four-step inclusion chain is given. We must avoid reversing inclusions and be careful with existential claims.Given Data / Assumptions:Clerk ⊆ Superintendent ⊆ Manager ⊆ Supervisor.Concept / Approach:From the chain, every Clerk is a Supervisor. If at least one Clerk exists (standard assumption in such tests), then “Some clerks are supervisors” follows. Other offered conclusions either reverse inclusions (I, IV) or require existence phrased differently (III) and are not included in the option key provided.Step-by-Step:1) Compose inclusions to get Clerk ⊆ Supervisor.2) Therefore, if Clerks exist, ∃ Clerk ∩ Supervisor (II).3) I would require Supervisor ⊆ Clerk (false).4) III is true under existence but is not provided as the keyed single-correct choice here; the best-supported single option is II.5) IV reverses inclusion and is false.Final Answer:Only conclusion II.

Question 18

Syllogism — Schools, laboratories, and theatres:  Statements: (a) Some schools are Laboratories. (b) All Laboratories are theatres.  Conclusions: I. Some theatres are schools. II. No school is a theatre.

Accepted Answer

Introduction / Context:We track a particular group through a universal inclusion.Given Data / Assumptions:Some Schools are Laboratories.All Laboratories are Theatres.Concept / Approach:Those Schools that are Laboratories must also be Theatres, so an overlap between Theatres and Schools is guaranteed. The universal negative in II contradicts this.Step-by-Step:1) Take the Schools that are Laboratories (exist).2) Map via Laboratory ⊆ Theatre to conclude they are Theatres.3) Therefore, ∃ (School ∩ Theatre): I follows and II is false.Final Answer:Conclusion I follows.

Question 19

Syllogism — Boys, men, and black (complements):  Statements: • Some boys are men. • No man is black.  Conclusions: I. Some boys are not black. II. Some men are boys.

Accepted Answer

Introduction / Context:We combine a particular inclusion with a universal negative to infer a particular negative and recognize an equivalent restatement.Given Data / Assumptions:Some Boys are Men (∃ Boy ∩ Man).No Man is Black (Man ⊆ not Black).Concept / Approach:The Boys who are Men are not Black, establishing I. Since “Some Boys are Men” is equivalent to “Some Men are Boys,” II also follows.Step-by-Step:1) From ∃ (Boy ∩ Man) and Man ⊆ (not Black), infer ∃ (Boy ∩ not Black) — that is, Some boys are not black (I).2) The same premise directly gives ∃ (Man ∩ Boy), which is II.Why Other Options Are Wrong:Any option excluding either I or II ignores necessary consequences.Final Answer:Both I and II follow.

Question 20

Syllogism — Teachers and bicycles; identifying what necessarily follows:  Statements: (a) No teacher comes to the school on a bicycle. (b) Anand comes to the school on a bicycle.  Conclusions: I. Anand is not a teacher. II. Anand is a student.

Accepted Answer

Introduction / Context:We apply a universal negative to a specific instance to see what must be excluded, avoiding unwarranted identity claims.Given Data / Assumptions:No Teacher is a Bicycle-Comer (to school).Anand is a Bicycle-Comer (to school).Concept / Approach:From “No T are B” and “Anand is B,” it follows “Anand is not T.” Nothing is stated about Anand being a student; he could be a non-teaching staff member, parent, visitor, etc.Step-by-Step:1) Translate: Teacher ∩ Bicycle = ∅.2) Anand ∈ Bicycle.3) Therefore, Anand ∉ Teacher (I follows).4) II is not supported by any premise.Final Answer:Conclusion I alone can be drawn.

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