Question 1

Statements: • All desks are chairs. • All chairs are tables. • All tables are boxes. • All boxes are trunks. Conclusions: I. Some trunks are tables. II. All chairs are boxes. III. Some boxes are desks. IV. All desks are trunks. Choose the option tha…

Accepted Answer

Introduction / Context:Pure subset chains allow safe transitive conclusions with universals, but do not create existence. Therefore, “some” conclusions require caution unless an existential premise is present.Given Data / Assumptions:Desks ⊆ Chairs ⊆ Tables ⊆ Boxes ⊆ Trunks.Concept / Approach:From the chain, any Chair is a Box (II true), and any Desk is a Trunk (IV true). However, “Some trunks are tables” and “Some boxes are desks” assert existence of Tables and Desks, which the premises do not provide.Step-by-Step Solution:• II: Chairs ⊆ Boxes by transitivity; universally true.• IV: Desks ⊆ Trunks by transitivity; universally true.• I: “Some trunks are tables” needs at least one Table; existence is not implied by the universals.• III: “Some boxes are desks” needs existence of a Desk; also not implied.Verification / Alternative check:Let all sets be empty; the universals remain true, yet I and III are false due to lack of witnesses.Why Other Options Are Wrong:They include existential claims not guaranteed by the premises.Common Pitfalls:Confusing universals with existence claims.Final Answer:Only II and IV follow.

Question 2

Statements: • Some birds are goats. • Some goats are horses. • Some horses are lions. • Some lions are tigers. Conclusions: I. Some tigers are goats. II. No tiger is a goat. III. Some lions are birds. IV. No lion is a bird. Choose the option that mu…

Accepted Answer

Introduction / Context:This is a chain of four independent “some” statements. Without any universal inclusion connecting them, none of the extreme overlaps is forced, and neither of the “either … or …” patterns is logically necessary.Given Data / Assumptions:Each premise provides a separate intersection: Birds∩Goats, Goats∩Horses, Horses∩Lions, Lions∩Tigers. They may involve entirely different individuals.Concept / Approach:To prove I (“Some tigers are goats”) we would need one element to travel through Goats, Horses, Lions into Tigers. The premises do not require that. To prove II (“No tiger is a goat”) we would need a universal exclusion between Tigers and Goats, which we do not have. The same analysis applies to III and IV with Lions and Birds.Step-by-Step Solution:• Build Model A where all four intersections are witnessed by different individuals; then I–IV are all false.• Build Model B where the chain coincides for a single individual; then I and III are true, II and IV are false. Because truth values vary with permitted models, none of I–IV is necessary.Verification / Alternative check:The “either … or … follows” options are used only when exactly one of a pair must be true. Here, both can be false in Model A, so neither “either … or …” is guaranteed.Why Other Options Are Wrong:They assume forced overlap or forced exclusion that is not present.Common Pitfalls:Confusing possibility with necessity in chains of “some.”Final Answer:None follows.

Question 3

Statements: • All papers are bottles. • All bottles are cups. • Some cups are jugs. • Some jugs are plates. Conclusions: I. Some plates are cups. II. Some plates are bottles. III. Some cups are papers. IV. Some bottles are papers. Choose the option …

Accepted Answer

Introduction / Context:Two universal inclusions are followed by two existential claims about later sets. We must be careful not to assert intersections that depend on the very same jug being both a cup and a plate.Given Data / Assumptions:Papers ⊆ Bottles ⊆ Cups.∃c_j ∈ Cups ∩ Jugs.∃j_p ∈ Jugs ∩ Plates.Concept / Approach:I and II require that the Cup-Jug element c_j is also a Plate (or the Jug-Plate element j_p is also a Cup). The premises do not force that identity. III and IV require existence of Papers; universal statements do not provide an existential witness.Step-by-Step Solution:• I–II: Not guaranteed because c_j and j_p can differ.• III–IV: “Some X are Papers” needs Papers to exist; not given.Verification / Alternative check:Let Papers be empty; let c_j and j_p be distinct. All premises remain true and I–IV fail.Why Other Options Are Wrong:They assume identity of witnesses or existential import not supplied.Common Pitfalls:Creating unintended chains across separate “some” statements.Final Answer:None follows.

Question 4

Statements: • All bulbs are wires. • No wire is a cable. • Some cables are brushes. • All brushes are paints. Conclusions: I. Some paints are cables. II. Some wires are bulbs. III. Some brushes are wires. IV. Some cables are bulbs. Choose the option…

Accepted Answer

Introduction / Context:This question combines a universal inclusion, a universal exclusion, and an existential that is pushed forward by another universal. We check which intersections are compelled.Given Data / Assumptions:Bulbs ⊆ Wires.Wires ∩ Cables = ∅.∃c_b ∈ Cables ∩ Brushes.Brushes ⊆ Paints.Concept / Approach:From c_b ∈ Brushes and Brushes ⊆ Paints, we obtain Paints ∩ Cables ≠ ∅ (I true). II requires existence of Bulbs, which is not given by “All bulbs are wires.” III and IV contradict or overreach given the exclusion: brushes need not be wires, and bulbs cannot be cables because bulbs are wires and wires are disjoint from cables.Step-by-Step Solution:• I: Guaranteed by pushing c_b through the universal inclusion to Paints.• II: “Some wires are bulbs” needs a bulb to exist; not stated.• III: No link wires↔brushes is provided.• IV: Impossible because Bulbs ⊆ Wires and Wires ∩ Cables = ∅.Verification / Alternative check:Let there be cables that are brushes; let there be no bulbs at all. Premises hold; I is true; II–IV fail.Why Other Options Are Wrong:They add existence or overlap that is not supported, or contradict the exclusion.Common Pitfalls:Assuming existence from a universal statement.Final Answer:Only I follows.

Question 5

Statements: • All chairs are keys. • All keys are balloons. • Some balloons are mirrors. • Some mirrors are desks. Conclusions: I. Some desks are keys. II. Some balloons are mirrors. III. Some mirrors are balloons. Choose the option that must follow.

Accepted Answer

Introduction / Context:Two universal inclusions are followed by two existentials forming a direct equivalence. We check which consequences are guaranteed without assuming that the desk-mirror items are among the keys.Given Data / Assumptions:Chairs ⊆ Keys ⊆ Balloons.∃b_m ∈ Balloons ∩ Mirrors.∃m_d ∈ Mirrors ∩ Desks.Concept / Approach:II and III are the same overlap phrased both ways and are explicitly given by the premise “Some balloons are mirrors.” I would need that some Desk is also a Key via Mirrors, which is not required: not all Balloons are Keys, so a Mirror can be a Balloon without being a Key.Step-by-Step Solution:• II and III: Guaranteed by the existential premise.• I: Would require a Mirror that is both a Desk and a Key. The premises do not force that identity.Verification / Alternative check:Model the Balloons∩Mirrors element outside Keys; then II–III remain true while I fails.Why Other Options Are Wrong:They include I, which assumes a non-forced overlap.Common Pitfalls:Assuming mirrors that are balloons are also keys; that is not assured.Final Answer:Only II and III follow.

Question 6

Statements: • Some drums are posters. • All posters are windows. • Some windows are tablets. • All tablets are books. Conclusions: I. Some windows are drums. II. Some books are posters. III. Some tablets are drums. Choose the option that must follow.

Accepted Answer

Introduction / Context:A universal inclusion pushes an existential backward to establish one guaranteed overlap. Other proposed overlaps require identity across independent “some” claims, which is not imposed by the premises.Given Data / Assumptions:∃d_p ∈ Drums ∩ Posters.Posters ⊆ Windows.∃w_t ∈ Windows ∩ Tablets.Tablets ⊆ Books.Concept / Approach:From d_p ∈ Posters and Posters ⊆ Windows, it follows that Windows ∩ Drums ≠ ∅ (I true). II would require Posters to reach Books via Tablets; not forced. III would need the Windows that are Tablets to coincide with the Windows that are Posters∩Drums; again, not forced.Step-by-Step Solution:• I: Direct consequence of the universal inclusion.• II: Needs some Poster to be a Tablet; not stated.• III: Needs the Tablet-window to be the Poster-window-drum; not stated.Verification / Alternative check:Choose disjoint witnesses for Windows∩Tablets and Drums∩Posters; then only I holds.Why Other Options Are Wrong:They add non-forced identities across different “some” statements.Common Pitfalls:Assuming all given windows are the same items.Final Answer:Only I follows.

Question 7

Statements: • Some trains are cars. • All cars are branches. • All branches are nets. • Some nets are dresses. Conclusions: I. Some dresses are cars. II. Some nets are trains. III. Some branches are trains. IV. Some dresses are trains. Choose the op…

Accepted Answer

Introduction / Context:A pair of universal inclusions push an existential from Trains to Nets. Another existential about Nets and Dresses does not necessarily meet the same elements, so conclusions involving Dresses and Trains or Cars are not guaranteed.Given Data / Assumptions:∃t_c ∈ Trains ∩ Cars.Cars ⊆ Branches ⊆ Nets.∃n_d ∈ Nets ∩ Dresses.Concept / Approach:From t_c and the universal steps, t_c ∈ Nets and t_c ∈ Branches, which guarantees II and III. Conclusions I and IV would require the Dress-Nets witness to be the same as t_c; not forced.Step-by-Step Solution:• Push t_c: Cars → Branches → Nets. Thus Nets ∩ Trains ≠ ∅ (II) and Branches ∩ Trains ≠ ∅ (III).• I and IV: Need overlap with Dresses; premises do not assert it.Verification / Alternative check:Set the Dresses∩Nets element distinct from t_c; II and III stay true; I and IV fail.Why Other Options Are Wrong:They include non-forced intersections with Dresses.Common Pitfalls:Assuming all “net” elements are identical across statements.Final Answer:Only II and III follow.

Question 8

Statements: • Some pencils are kites. • Some kites are desks. • All desks are jungles. • All jungles are mountains. Conclusions: I. Some mountains are pencils. II. Some jungles are pencils. III. Some mountains are desks. IV. Some jungles are kites. …

Accepted Answer

Introduction / Context:A universal ladder from Desks up to Mountains lets us promote existence from “Some kites are desks.” We test which conclusions are forced by that promotion and which would require extra overlaps.Given Data / Assumptions:∃k_d ∈ Kites ∩ Desks.Desks ⊆ Jungles ⊆ Mountains.∃p_k ∈ Pencils ∩ Kites (may be different from k_d).Concept / Approach:From k_d and the universal steps, k_d ∈ Jungles and k_d ∈ Mountains, so III (“Some mountains are desks”) and IV (“Some jungles are kites”) are guaranteed. I and II require that the Pencil-Kite witness is the same as k_d; not required.Step-by-Step Solution:• Push k_d up: Desks → Jungles → Mountains.• Conclude III and IV immediately.• Note that nothing forces Pencils to overlap those Desks/Jungles, so I and II do not necessarily follow.Verification / Alternative check:Let the Pencil-Kite element lie outside Desks. Then III and IV remain true, while I and II fail.Why Other Options Are Wrong:They include non-forced intersections with Pencils.Common Pitfalls:Assuming all kites in different premises refer to the same items.Final Answer:Only III and IV follow.

Question 9

Statements: • All papers are clips. • Some chips are boards. • Some boards are lanes. • All lanes are roads. Conclusions: I. Some roads are boards. II. Some lanes are chips. III. Some boards are papers. IV. Some roads are chips. Choose the option th…

Accepted Answer

Introduction / Context:A universal inclusion takes “Some boards are lanes” directly to roads. Other proposed overlaps need identities across different “some” statements or links to Papers/Clips that are not provided.Given Data / Assumptions:∃b_l ∈ Boards ∩ Lanes.Lanes ⊆ Roads.∃c_b ∈ Chips ∩ Boards (may be different from b_l).Papers ⊆ Clips (no connection to Boards given).Concept / Approach:From b_l and Lanes ⊆ Roads, b_l ∈ Roads ∩ Boards, hence I is guaranteed. II and IV would require that the same board is both a lane and a chip; not forced. III would require an overlap between Boards and Papers; not given.Step-by-Step Solution:• I: Directly from pushing b_l through Lanes ⊆ Roads.• II–IV: Depend on unforced identities or missing links; they do not necessarily follow.Verification / Alternative check:Let b_l and c_b be distinct; connect Papers only to Clips. Then only I holds.Why Other Options Are Wrong:They assert overlaps not entailed by the premises.Common Pitfalls:Assuming all “board” elements coincide across premises.Final Answer:Only I follows.

Question 10

Statements: • All pens are clocks. • Some clocks are tyres. • Some tyres are wheels. • Some wheels are buses. Conclusions: I. Some buses are tyres. II. Some wheels are clocks. III. Some wheels are pens. IV. Some buses are clocks. Choose the option t…

Accepted Answer

Introduction / Context:This is a chain of three “some” statements with a single unrelated universal. Without a universal link across those “some” statements, no intersection among the endpoints is forced.Given Data / Assumptions:∃c_t ∈ Clocks ∩ Tyres; ∃t_w ∈ Tyres ∩ Wheels; ∃w_b ∈ Wheels ∩ Buses; Pens ⊆ Clocks. The witnesses can all be different.Concept / Approach:I needs the Wheel-Bus and Tyre-Wheel witnesses to coincide; not forced. II needs the Tyre-Clock to also be a Wheel; not forced. III then needs that Wheel to be a Pen; not forced. IV needs a Wheel-Bus that is also a Clock; not forced.Step-by-Step Solution:• Construct a model with three distinct witnesses c_t, t_w, w_b. All premises hold; I–IV are all false.• Because truth values can vary across permitted models, none of the conclusions is necessary.Verification / Alternative check:Even though Pens ⊆ Clocks, there is no statement connecting Pens to Tyres or Wheels.Why Other Options Are Wrong:They overstate what “some” statements can guarantee.Common Pitfalls:Assuming transitivity for “some” across multiple steps.Final Answer:None follows.

Question 11

Statements: • All stones are hammers. • No hammer is a ring. • Some rings are doors. • All doors are windows. Conclusions: I. Some windows are stones. II. Some windows are rings. III. No window is a stone. IV. Some rings are stones. Choose the optio…

Accepted Answer

Introduction / Context:Universal exclusion (no hammer is a ring) coexists with a universal inclusion (doors ⊆ windows) and an existential about rings and doors. We look for the conclusion that is compelled regardless of how other sets relate.Given Data / Assumptions:Stones ⊆ Hammers.Hammers ∩ Rings = ∅.∃r_d ∈ Rings ∩ Doors.Doors ⊆ Windows.Concept / Approach:From r_d ∈ Doors and Doors ⊆ Windows, r_d ∈ Windows, so Windows ∩ Rings ≠ ∅ (II true). I would need Stones to reach Windows via Doors or another link; not given. III (“No window is a stone”) is not implied, because nothing forbids some Window from being a Hammer (and thus possibly a Stone) apart from ring-related exclusions. IV contradicts the exclusion because Stones ⊆ Hammers and Hammers ∩ Rings = ∅.Step-by-Step Solution:• Push r_d through Doors ⊆ Windows to prove II.• Note that there is no route forcing Stones into Windows; I does not follow.• III is a universal negative about Windows and Stones with no supporting premise.• IV is impossible due to Hammers ∩ Rings = ∅ and Stones ⊆ Hammers.Verification / Alternative check:Model Windows as a large set containing Doors plus other items; allow some Windows to overlap Hammers without being Rings. Premises remain consistent; only II is guaranteed.Why Other Options Are Wrong:They add claims not enforced by the premises or contradict them.Common Pitfalls:Reading ring-related exclusions as forbidding all other overlaps with Windows.Final Answer:Only II follows.

Question 12

Syllogism – Choose the necessary conclusion from the given statements: Statements: 1) All students in my class are bright. 2) Sarla is not bright. Conclusions: I) All students in my class are bright. II) Sarla is not bright. Which alternative states…

Accepted Answer

Introduction / Context:This is a classic categorical syllogism about set membership and complements. We are told every student in the speaker’s class belongs to the set Bright, and that Sarla is outside Bright. We must infer Sarla’s relation to the set of students in my class.Given Data / Assumptions:Let Class denote the set of “students in my class”.Premise 1: Class ⊆ Bright.Premise 2: Sarla ∉ Bright.No other information about “work hard”, or the population of non-bright persons, is provided.Concept / Approach:If every member of Class is Bright, then any non-Bright individual cannot be in Class (contrapositive reasoning on subset membership). This yields a definite conclusion about Sarla’s membership.Step-by-Step Solution:From Class ⊆ Bright, membership in Class implies membership in Bright.Given Sarla ∉ Bright, she cannot satisfy the implication’s antecedent.Therefore Sarla ∉ Class.Verification / Alternative check:Construct a model with Class = {x,y}, Bright = {x,y,z}, Sarla = s, and s ∉ Bright. All students in Class are bright, but Sarla is not bright; hence Sarla is outside Class. The reasoning is model-independent.Why Other Options Are Wrong:“Sarla must work hard”: irrelevant; “hard work” is not in the premises.“Some students are not bright”: contradicts Premise 1.“None-bright ones are not students”: over-generalizes beyond the specific class referenced; we do not talk about all students everywhere.Common Pitfalls:Assuming statements about an entire universe of “students” when the premise restricts to “my class”. Also, confusing necessity with probability.Final Answer:Sarla is not a student of my class.

Question 13

Assumption Evaluation – Decide which assumptions are implicit in the statements: Statements: (a) All girls do not like to marry (i.e., no girl likes to marry). (b) Some girls do not marry. Assumptions to test: I) All girls like to marry. II) All gir…

Accepted Answer

Introduction / Context:This problem asks which assumptions are implicitly taken for granted by the given statements. An assumption is something the statement relies on being true, not a conclusion we can forcibly derive.Given Data / Assumptions (from the stem):(a) Interprets as “No girl likes to marry.”(b) “Some girls do not marry.”Assumption I: “All girls like to marry.”Assumption II: “All girls do not marry.”Concept / Approach:Implicit assumptions should support or be required by the statements. Any assumption contradicting a statement is invalid; any assumption that is stronger than what is needed (and not required) is also invalid.Step-by-Step Solution:Assumption I directly contradicts (a), which declares no girl likes to marry. Hence I cannot be implicit.Assumption II is stronger than (b). Statement (b) asserts an existential claim (“some”), not a universal (“all”). The statement does not require that every girl fails to marry; therefore II is not implicit.Verification / Alternative check:A consistent model exists where some girls marry, some do not, and (a) is simply a (controversial) claim about liking; regardless, (b) needs only “at least one girl does not marry”. Neither I nor II is required for the statements to make sense.Why Other Options Are Wrong:I is valid / II is valid / Both valid: each accepts an assumption that either contradicts (I) or overstates necessity (II).Common Pitfalls:Confusing “implicit assumption” with “stronger version that would also make the statement true.” Implicit means necessary, not merely sufficient.Final Answer:Both I and II are invalid.

Question 14

Syllogism – Determine which conclusions necessarily follow: Statements: (I) All cities are towns. (II) Some cities are villages. Conclusions: (I) All villages are towns. (II) No village is a town. (III) Some villages are towns.

Accepted Answer

Introduction / Context:We must test which conclusion(s) are forced by the premises, true in every model that satisfies them. Universals give subset relations; particulars give existence.Given Data / Assumptions:All Cities are Towns (C ⊆ T).Some Cities are Villages (∃ C ∩ V).Concept / Approach:If C ⊆ T and some C are V, then those particular C that are V are also in T. Therefore there exists at least one V that is a T, which is exactly conclusion (III). Conclusions (I) and (II) are universal negatives/positives and are far too strong.Step-by-Step Solution:Take x ∈ C ∩ V (exists by statement II).Since C ⊆ T, x ∈ T as well.Therefore x ∈ V ∩ T, proving ∃ V ∩ T, i.e., “Some villages are towns.”Verification / Alternative check:Build a diagram with C entirely inside T, and let C overlap V in a small region. That overlap sits inside T, so at least one Village is a Town, but not necessarily all Villages are Towns, nor necessarily none.Why Other Options Are Wrong:(I) “All villages are towns” is not forced; V can extend outside T.(II) “No village is a town” contradicts the demonstrable overlap.“None of these” fails because (III) does follow.Common Pitfalls:Confusing “some” with “all,” or assuming disjointness without evidence.Final Answer:Only conclusion (III) follows.

Question 15

Logical consistency – Which of the following statements is certainly true (independent claims)? I. “All players are not tall” (i.e., not all players are tall). II. All basketball players are tall. III. All tall people are players. IV. Some players a…

Accepted Answer

Introduction / Context:The task is to identify which single claim can be accepted as universally valid without depending on the truth of the others. Items I–III are universal or quasi-universal claims and are easy to defeat by counterexample; item IV is existential and aligns with general real-world plausibility in aptitude settings.Given Data / Assumptions:The items are independent statements; we are not to combine them as premises.“All players are not tall” is typically read in tests as “Not all players are tall”.Concept / Approach:To be “certainly true,” a statement must hold across reasonable interpretations and avoid over-generalization. Claims II and III are sweeping universals and can be false in many natural models. Claim I is a particular style ambiguity and is not guaranteed. Claim IV (“Some players are tall”) is mild and matches everyday knowledge, and is the only one commonly accepted as “true” in such question forms.Step-by-Step Solution:Provide a counterexample to II: a short basketball player would falsify it.Provide a counterexample to III: a tall non-player falsifies it.Ambiguity in I prevents certainty.IV states an existential fact consistent with the domain: there exist tall players; this is overwhelmingly safe and intended as the test key.Verification / Alternative check:In standardized syllogism practice, when faced with one natural existential and several sweeping universals lacking support, the existential is chosen as the certain truth.Why Other Options Are Wrong:They assert universals without premise support and are easily refuted.Common Pitfalls:Reading I as “All players are tall” (the opposite) or treating the set of tall people as identical to players.Final Answer:IV.

Question 16

Club membership & exclusion – Infer the necessary conclusion: Statements: • All members of the Tennis club are also members of the Badminton club. • No woman plays Badminton. Which option must be true?

Accepted Answer

Introduction / Context:This question combines a subset relation with an exclusion statement. If Tennis ⊆ Badminton and no woman plays Badminton, then women are excluded from every subset of Badminton, including the Tennis club.Given Data / Assumptions:Tennis ⊆ Badminton.No Woman ∈ Badminton.Concept / Approach:Exclusion propagates downward through subsets: if a group is barred from a superset, it is barred from all its subsets. Therefore the women-exclusion at Badminton level automatically blocks membership in Tennis.Step-by-Step Solution:Assume for contradiction that some woman is a Tennis member.Because Tennis ⊆ Badminton, she would then be a Badminton player.This contradicts “No woman plays Badminton.” Thus no woman can be a Tennis member.Verification / Alternative check:A Venn diagram with Tennis circle entirely inside Badminton and with the entire Badminton circle labeled “no women” shows Tennis shares the same exclusion.Why Other Options Are Wrong:“Some women play Tennis” / “Some women are Tennis members”: impossible by the exclusion.“No Tennis member plays Badminton”: false; in fact every Tennis member is also a Badminton member.Common Pitfalls:Confusing “subset” with “disjoint,” or reading the second statement as “no woman is a Badminton club member” in a weaker, non-playing sense (the logic here is about playing/membership as stated).Final Answer:No woman is a member of the Tennis club.

Question 17

Syllogism – Decide which conclusion(s) are forced: Statements: • Some Indians are educated. • Educated men prefer small families. Conclusions: (I) All small families are educated. (II) Some Indians prefer small families.

Accepted Answer

Introduction / Context:This problem checks careful reading of quantifiers and subgroup qualifiers. We have one existential about “Indians” and one conditional preference limited to “educated men”.Given Data / Assumptions:Some Indians are educated (∃ Indian ∩ Educated).Educated men ⊆ PreferSmallFamilies.No statement guarantees that the “some educated Indians” are men, nor that every small family is connected back to “educated”.Concept / Approach:(I) reverses the given direction (“educated men ⇒ prefer small families”) into a universal about all small families; this is a converse fallacy. (II) needs at least one Indian who is both educated and a man; “Some Indians are educated” does not require that the “some” are men.Step-by-Step Solution:To force (II), we would need ∃ (Indian ∩ Educated ∩ Man). The premises do not assert this.To force (I), we would need SmallFamily ⊆ Educated (or ⊆ EducatedMen), which is absent.Verification / Alternative check:Countermodel: All educated Indians are women; educated men (if any) are non-Indians. Then (II) fails. Also, there can be small families formed by uneducated persons, so (I) fails.Why Other Options Are Wrong:Any option claiming a forced conclusion ignores the missing links (gender condition for II; converse direction for I).Common Pitfalls:Forgetting subgroup qualifiers (“men”) and illicitly converting conditionals.Final Answer:Neither I nor II follows.

Question 18

Direct syllogism – Identify the necessary conclusion: Statements: • All men are mortal. • Ramu is a man. Which conclusion must follow?

Accepted Answer

Introduction / Context:A textbook categorical syllogism: from a universal inclusion to a specific instance.Given Data / Assumptions:Men ⊆ Mortal.Ramu ∈ Men.Concept / Approach:Membership in a subset compels membership in the superset. Therefore the individual inherits the property “mortal”.Step-by-Step Solution:From Men ⊆ Mortal and Ramu ∈ Men, infer Ramu ∈ Mortal.Verification / Alternative check:No countermodel exists because the universal premise makes it impossible for any man to be outside Mortal.Why Other Options Are Wrong:They contradict the universal premise.Common Pitfalls:None here; this is the archetype of correct syllogistic instantiation.Final Answer:Ramu is mortal.

Question 19

Syllogism – Shared superset without guaranteed overlap: Statements: (a) All poets are intelligent. (b) All singers are intelligent. Conclusions: I) All singers are poets. II) Some intelligent persons are not singers.

Accepted Answer

Introduction / Context:Two different groups lie inside the same superset (Intelligent). The trap is to infer overlap where none is stated.Given Data / Assumptions:Poet ⊆ Intelligent.Singer ⊆ Intelligent.No premise about Poet–Singer intersection or about the size of Intelligent.Concept / Approach:(I) “All singers are poets” is unjustified: two distinct subsets of a superset can be disjoint. (II) “Some intelligent persons are not singers” is also not compelled: it is possible (though unlikely in reality) that every intelligent person is a singer in some models. Syllogism evaluates necessity across all models, not plausibility.Step-by-Step Solution:Construct Model A: Poets and Singers are disjoint proper subsets of Intelligent. (I) fails; (II) holds here but is not necessary.Construct Model B: Intelligent = Singers, with Poets ⊆ Singers. (I) may still fail, but (II) fails because every Intelligent is a Singer. Because conclusions can flip, neither is necessary.Verification / Alternative check:Necessity requires truth in all valid models. Since countermodels exist for both I and II, neither follows.Why Other Options Are Wrong:They claim necessity where only possibility exists.Common Pitfalls:Assuming “shared superset” implies overlap or non-overlap without explicit statements.Final Answer:Neither conclusion I nor II follows.

Question 20

Implication check – What can be concluded from “All beggars are poor”? Assume standard category meanings for “beggar”, “poor”, and “rich”.

Accepted Answer

Introduction / Context:We test typical aptitude-exam reading of socioeconomic labels: “All beggars are poor” is taken with the everyday background that “rich” and “poor” are mutually exclusive. Within that convention, the question asks for a safe implication.Given Data / Assumptions:Premise: Beggar ⊆ Poor.Background convention in such questions: Poor ∩ Rich = ∅ (disjoint economic categories).Concept / Approach:From Beggar ⊆ Poor and Poor ∩ Rich = ∅, it follows that Beggar ∩ Rich = ∅. Hence any specific beggar cannot be rich. Be careful not to invert the subset (which would claim all poor are beggars).Step-by-Step Solution:Take any individual x. If x is a Beggar, then x is Poor (by premise).If x is Poor, then x is not Rich (by background disjointness).Therefore, if x is a Beggar, x is not Rich.Verification / Alternative check:Venn diagram: Beggar region sits entirely inside Poor; Rich is disjoint from Poor. Hence Beggar and Rich do not overlap.Why Other Options Are Wrong:“All poor are beggars” reverses the subset.“If A is rich then not a beggar” requires Rich ⇒ not Poor; while commonly assumed, the logically safest choice focused on the given premise is the beggar-focused conditional (option D).“If A is not rich then not a beggar” is invalid; non-rich includes poor and middle-income, and could include beggars.Common Pitfalls:Converse and inverse errors when handling subset statements.Final Answer:If A is a beggar, then A is not rich.

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