Question 1

Syllogism (logical deduction): Evaluate the conclusions based on the given statements.  Statements: S1: Some actors are singers. S2: All the singers are dancers.  Conclusions: C1: Some actors are dancers. C2: No singer is actor.  Choose the option t…

Accepted Answer

Given data S1: Some actors are singers. S2: All singers are dancers. Conclusions to test: C1 (Some actors are dancers), C2 (No singer is actor). Concept / Approach Use set relations: If some Actors ∩ Singers ≠ ∅ and Singers ⊆ Dancers, then those same actors are also Dancers. A universal negative about singers vs actors contradicts S1. Step-by-step check From S1: ∃x such that x ∈ Actors and x ∈ Singers. From S2: Singers ⊆ Dancers ⇒ Any singer is a dancer. Therefore that x is also a Dancer ⇒ Some actors are dancers ⇒ C1 is true. C2 says “No singer is actor,” which directly contradicts S1 ⇒ C2 is false. Common pitfalls Assuming “some” means “all,” or overlooking subset propagation (Singers ⊆ Dancers). Final Answer Only conclusion I follows.

Question 2

Syllogism (set inclusion): Decide which conclusion(s) logically follow from the statements.  Statements: S1: All the harmoniums are instruments. S2: All the instruments are flutes.  Conclusions: C1: All the flutes are instruments. C2: All the harmon…

Accepted Answer

Given data S1: Harmoniums ⊆ Instruments. S2: Instruments ⊆ Flutes. Conclusions: C1 “All flutes are instruments,” C2 “All harmoniums are flutes.” Concept / Approach Subset chaining: If A ⊆ B and B ⊆ C, then A ⊆ C. Converse statements are not guaranteed. Testing conclusions From S1 and S2: Harmoniums ⊆ Instruments ⊆ Flutes ⇒ Harmoniums ⊆ Flutes ⇒ C2 is true. C1 is the converse of S2 (claims Flutes ⊆ Instruments). S2 only gives Instruments ⊆ Flutes, not the reverse ⇒ C1 is not guaranteed. Final Answer Only conclusion II follows.

Question 3

Syllogism (categorical reasoning): Determine which conclusions follow strictly.  Statements: S1: Some mangoes are yellow. S2: Some tixo are mangoes.  Conclusions: C1: Some mangoes are green. C2: Tixo is a yellow.  Choose the correct option according…

Accepted Answer

Given data S1: Some Mangoes are Yellow (existential statement about Yellow mangoes). S2: Some Tixo are Mangoes (existential statement linking Tixo to Mangoes). Concept / Approach “Some” statements do not force overlap with other “Some” statements unless explicitly connected. No information is given about “green,” and there is no necessary overlap between the specific yellow mangoes of S1 and the specific mangoes related to Tixo in S2. Testing conclusions C1: “Some mangoes are green” is unsupported; color “green” is never mentioned. C2: From S1 and S2, we cannot conclude Tixo are Yellow, because the “some mangoes” related to Tixo need not be the same “some mangoes” that are Yellow. Final Answer Neither I nor II follows.

Question 4

Syllogism (set chaining): Identify which conclusions follow from the premises.  Statements: S1: Some ants are parrots. S2: All the parrots are apples.  Conclusions: C1: All the apples are parrots. C2: Some ants are apples.  Choose the correct option.

Accepted Answer

Given data S1: ∃x such that x ∈ Ants and x ∈ Parrots. S2: Parrots ⊆ Apples. Concept / Approach When some Ants are Parrots and every Parrot is an Apple, those same Ants are Apples. The converse universal claim about Apples need not hold. Testing conclusions C2: True, because the intersecting elements from S1 are inside Parrots, and all Parrots are Apples ⇒ Some Ants are Apples. C1: False, because S2 gives Parrots ⊆ Apples, not Apples ⊆ Parrots. Final Answer Only conclusion II follows.

Question 5

Syllogism (pens, pencils, papers): Determine which conclusions must be true.  Statements: S1: Some papers are pens. S2: All the pencils are pens.  Conclusions: C1: Some pens are pencils. C2: Some pens are papers.  Choose the correct option.

Accepted Answer

Given data S1: Some Papers ∩ Pens ≠ ∅. S2: Pencils ⊆ Pens. Concept / Approach Universal inclusion does not imply existence. From S2 alone, we cannot assert that any Pencil exists to guarantee “Some pens are pencils.” But S1 already guarantees at least one Pen that is also a Paper. Testing conclusions C1: Not guaranteed; there may be no Pencils at all. “All Pencils are Pens” does not ensure “Some Pens are Pencils.” C2: Guaranteed by S1, because if some Papers are Pens, then some Pens are Papers (same intersection). Final Answer Only conclusion II follows.

Question 6

Syllogism (actors, girls, beautiful): Evaluate the validity of the conclusions.  Statements: S1: All the actors are girls. S2: All the girls are beautiful.  Conclusions: C1: All the actors are beautiful. C2: Some girls are actors.  Choose the correc…

Accepted Answer

Given data S1: Actors ⊆ Girls. S2: Girls ⊆ Beautiful. Concept / Approach Subset chaining: If Actors ⊆ Girls and Girls ⊆ Beautiful, then Actors ⊆ Beautiful. The statement “Some Girls are Actors” requires existential import for Actors, which is not guaranteed by universal statements alone. Testing conclusions C1: True by transitivity ⇒ Actors ⊆ Beautiful. C2: Not guaranteed; from universals alone, we cannot assert existence of Actors. Final Answer Only conclusion I follows.

Question 7

Syllogism (windows, doors, walls): Decide which conclusion(s) follow logically.  Statements: S1: All the windows are doors. S2: No door is a wall.  Conclusions: C1: Some windows are walls. C2: No wall is a door.  Choose the correct option.

Accepted Answer

Given data S1: Windows ⊆ Doors. S2: Doors ∩ Walls = ∅ (No door is a wall). Concept / Approach From S2, the disjointness is symmetric: If no A is B, then no B is A. Also, since Windows are a subset of Doors, Windows cannot overlap with Walls. Testing conclusions C1: “Some windows are walls” contradicts S1+S2, because any Window is a Door and no Door is a Wall. C2: “No wall is a door” is the contrapositive-symmetric form of S2, valid for disjointness statements. Final Answer Only conclusion II follows.

Question 8

Syllogism: Evaluate which conclusions logically follow from the given statements (Venn-diagram reasoning).  Statements: All cups are books. All books are shirts.  Conclusions: I. Some cups are not shirts. II. Some shirts are cups.  Choose the correc…

Accepted Answer

Given data Statement 1: All cups ⊆ books. Statement 2: All books ⊆ shirts. Therefore: All cups ⊆ shirts. Conclusions to test: (I) Some cups are not shirts; (II) Some shirts are cups. Concept/Approach Translate universal statements into set containment and test each conclusion against all possible Venn diagrams consistent with the premises. From All cups ⊆ books and All books ⊆ shirts, it follows that all cups are inside shirts. Step-by-step evaluation Conclusion I: "Some cups are not shirts" contradicts All cups ⊆ shirts, so it does not follow. Conclusion II: Since cups are a subset of shirts and we assume the existence of at least one cup (standard syllogism convention in such MCQs), at least one element is both a shirt and a cup; hence "Some shirts are cups" follows. Common pitfalls Confusing the converse ("Some shirts are cups") with the original universal; here, particular existence is acceptable under exam conventions. Final Answer Only Conclusion II follows

Question 9

Syllogism: Evaluate which conclusions logically follow from the given statements (possibility vs. necessity).  Statements: Some cows are crows. Some crows are elephants.  Conclusions: I. Some cows are elephants. II. All crows are elephants.  Choose …

Accepted Answer

Given data Some cows ∩ crows ≠ ∅. Some crows ∩ elephants ≠ ∅. Conclusions to test: (I) Some cows ∩ elephants ≠ ∅; (II) All crows ⊆ elephants. Concept/Approach "Some–Some" premises do not force an overlap between the first and third sets; the overlapping parts of crows could be disjoint (different crows in each statement). Step-by-step evaluation Conclusion I: It is only a possibility, not a necessity. There exist valid diagrams where cows and elephants do not overlap, so it does not follow. Conclusion II: "All crows are elephants" is far stronger than the premises and does not follow. Final Answer Neither Conclusion I nor II follows

Question 10

Syllogism: Determine which conclusions must follow from the chain of containments.  Statements: All the pencils are pens. All the pens are inks.  Conclusions: I. All the pencils are inks. II. Some inks are pencils.  Choose the correct option about w…

Accepted Answer

Given data All pencils ⊆ pens; All pens ⊆ inks ⇒ All pencils ⊆ inks. Conclusions: (I) All pencils are inks; (II) Some inks are pencils. Concept/Approach Transitivity of containment validates I. Under standard MCQ convention assuming existence of pencils, at least one pencil exists, so that element is an ink; hence "Some inks are pencils" holds. Step-by-step evaluation I: Directly true by transitivity (pencils ⊆ inks). II: From existence of at least one pencil in the subset, "some" particular of inks is indeed a pencil. Final Answer Both Conclusions I and II follow

Question 11

Syllogism: Assess whether the given pairwise "some" statements force an overlap between outer sets.  Statements: Some dogs are bats. Some bats are cats.  Conclusions: I. Some dogs are cats. II. Some cats are dogs.  Choose the correct option about wh…

Accepted Answer

Given data Some dogs overlap with bats; some bats overlap with cats. Middle term is "bats". Concept/Approach Two separate "some" overlaps via a middle set do not compel dogs and cats to overlap; the overlapping bats could be different individuals. Evaluation Both I and II are mere possibilities, not necessary conclusions. Hence neither follows. Final Answer Neither Conclusion I nor II follows

Question 12

Syllogism: Decide which conclusions necessarily follow from the category statements.  Statements: All the trucks are flies. Some scooters are flies.  Conclusions: I. All the trucks are scooters. II. Some scooters are trucks.  Choose the correct opti…

Accepted Answer

Given data All trucks ⊆ flies; Some scooters ∩ flies ≠ ∅. No direct relation between "trucks" and "scooters" is compelled. Analysis I: "All trucks are scooters" is not forced—trucks could be a different subset of flies. II: "Some scooters are trucks" is also not necessary—overlap may or may not exist. Final Answer Neither Conclusion I nor II follows

Question 13

Syllogism: Test the conclusions using set relations and complement information.  Statements: All buildings are chalks. No chalk is toffee.  Conclusions: I. No building is toffee. II. All chalks are buildings.  Choose the correct option about which c…

Accepted Answer

Given data All buildings ⊆ chalks. No chalk ∩ toffee ≠ ∅ (i.e., chalks and toffee are disjoint). Reasoning If every building is a chalk and chalks have no overlap with toffee, then buildings cannot be toffee. Thus I follows. "All chalks are buildings" is the converse of the first statement and does not follow. Final Answer Only Conclusion I follows

Question 14

Syllogism: Determine the logically valid conclusion(s) based on two subset statements with a common superser.  Statements: All cars are cats. All fans are cats.  Conclusions: I. All cars are fans. II. Some fans are cars.  Choose the correct option a…

Accepted Answer

Given data Cars ⊆ cats; Fans ⊆ cats. No direct relationship stated between cars and fans. Evaluation I: "All cars are fans" need not be true; both could be disjoint subsets of cats. II: "Some fans are cars" is only a possibility, not a necessity. Final Answer Neither Conclusion I nor II follows

Question 15

Syllogism — Categorical logic on 'door', 'dog', and 'cat' Statements: • No door is dog. • All the dogs are cats. Conclusions to evaluate:  No door is cat.  No cat is door.  Some cats are dogs.  All the cats are dogs.

Accepted Answer

Introduction / Context:This is a classic syllogism problem involving universal negatives and affirmatives. The goal is to determine which conclusions logically follow from the given statements without adding external assumptions beyond standard exam conventions.Given Data / Assumptions:Premise 1: 'No door is dog' (Door ∩ Dog = ∅).Premise 2: 'All dogs are cats' (Dog ⊆ Cat).Typical exam convention allows existential import for 'dogs' when inferring 'Some cats are dogs' from 'All dogs are cats' (i.e., dogs are assumed to exist).Concept / Approach:Use standard Venn logic. From Dog ⊆ Cat, every dog lies within the Cat circle. From No Door is Dog, the Door set is disjoint from Dog, but it might still overlap with the Cat region that is not Dog. Therefore, universal negatives about Door vs Cat do not necessarily follow.Step-by-Step Solution:From 'All dogs are cats': If at least one dog exists, then 'Some cats are dogs' is true → Conclusion (3) follows under common test assumptions.From 'No door is dog': We only know Door excludes the Dog subset. Cats outside Dog remain unrestricted regarding Door, so we cannot assert 'No door is cat' or its converse.There is no basis for 'All the cats are dogs' because Dog is a subset of Cat, not vice versa.Verification / Alternative check:Construct a model: Let Dog = {d1}, Cat = {d1, c2}, Door = {r1}. Then Premise 1 and 2 hold. Here, some cats are dogs (d1 ∈ Cat ∩ Dog) is true; but 'No door is cat' is not determined (r1 could be non-cat or cat); 'All cats are dogs' is false.Why Other Options Are Wrong:(1) and (2): Neither necessarily follows; Door could overlap Cat outside Dog.(4): Reverses the subset relation; not supported.Combinations including any of (1), (2), or (4) are therefore invalid.Common Pitfalls:Assuming that because doors are not dogs and dogs are cats, doors are not cats. This ignores the possibility of cats that are not dogs.Final Answer:Only (3)

Question 16

Logical syllogism — determine which conclusions follow beyond doubt.  Statements: 1) All green are blue. 2) All blue are white.  Conclusions: (1) Some blue are green. (2) Some white are green. (3) Some green are not white. (4) All white are blue.

Accepted Answer

Introduction / Context: This verbal reasoning item tests classic categorical syllogism skills using set relations such as “All A are B”. The goal is to decide which conclusions must follow from the given statements, not which could be true.Given Data / Assumptions: Statement 1: All green are blue.Statement 2: All blue are white.Standard test assumption: the classes mentioned are non-empty for “Some …” derivations in traditional syllogism questions.Concept / Approach: Chain rules: if all A are B and all B are C, then all A are C. Immediate inferences: from “All A are B”, we commonly accept “Some B are A” in these exam styles (existential import). Also, converse statements like “All C are B” generally do not follow.Step-by-Step Solution: From 1 and 2, derive: all green are white (green ⊆ blue ⊆ white). Test (1): “Some blue are green.” Since at least one green exists and every green is blue, some blue are indeed green. Conclusion (1) follows. Test (2): “Some white are green.” From green ⊆ white and existence of green, at least some white are green. Conclusion (2) follows. Test (3): “Some green are not white.” This contradicts green ⊆ white, so it does not follow. Test (4): “All white are blue.” This is the converse of statement 2 and is not implied. It may be false if some white are not blue.Verification / Alternative check: Think of circles: green inside blue, blue inside white. Visualizing confirms (1) and (2) must be true, while (3) and (4) need not be.Why Other Options Are Wrong: Options listing (3) are wrong since (3) contradicts the chain. Options listing (4) assume a converse which is not guaranteed. Hence only the option with (1) and (2) is valid.Common Pitfalls: Confusing “All B are C” with “All C are B” (converse error) and forgetting the typical exam convention that allows deriving a “Some …” statement from an “All …” statement given existence.Final Answer: Only (1) and (2)

Question 17

Logical syllogism — determine which conclusions follow beyond doubt.  Statements: 1) All men are vertebrates. 2) Some mammals are vertebrates.  Conclusions: (1) All men are mammals. (2) All mammals are men. (3) Some vertebrates are mammals. (4) All …

Accepted Answer

Introduction / Context: This problem checks your ability to recognize necessary conclusions from two categorical premises involving sets: men, mammals, and vertebrates.Given Data / Assumptions: All men ⊆ vertebrates.Some mammals ⊆ vertebrates.We must accept only what must be true in every model that satisfies the statements.Concept / Approach: From “Some mammals are vertebrates,” we can immediately convert to “Some vertebrates are mammals.” Universals (“All …”) do not allow converses (e.g., “All vertebrates are men”) without evidence. Also, nothing connects all men to mammals directly.Step-by-Step Solution: Check (3): Some vertebrates are mammals — directly supported by Statement 2 via simple conversion; hence it follows. Check (1): All men are mammals — not implied; men could be vertebrates that are not mammals. Check (2): All mammals are men — no support; mammals could include many non-human species. Check (4): All vertebrates are men — unjustified and clearly false in general.Verification / Alternative check: Construct an example: let vertebrates include fish and men; let mammals include dogs (a subset of vertebrates). Then “Some vertebrates are mammals” is true (dogs), but (1), (2), (4) are not forced. This validates only (3).Why Other Options Are Wrong: (1) and (2) over-generalize; (4) is an extreme converse/generalization without basis. Only (3) is logically compelled.Common Pitfalls: Mistaking subset relations for equality, and assuming converses of universal statements are true.Final Answer: Only (3)

Question 18

Logical syllogism — determine which conclusions follow beyond doubt.  Statements: 1) All the phones are scales. 2) All the scales are calculators.  Conclusions: (1) All the calculators are scales. (2) All the phones are calculators. (3) All the scal…

Accepted Answer

Introduction / Context: Two universal premises describe a chain: phones ⊆ scales ⊆ calculators. We must find the conclusions that necessarily follow from this chain of inclusion.Given Data / Assumptions: All phones are scales.All scales are calculators.Standard existence assumption for “Some …” based on presence of the subject class.Concept / Approach: Transitivity of subset: if A ⊆ B and B ⊆ C, then A ⊆ C. Converse statements such as “All C are B” do not follow. From “All A are C,” we may infer “Some C are A” when A is non-empty, which these tests typically assume.Step-by-Step Solution: From the chain, phones ⊆ calculators, so conclusion (2) holds. Because phones exist, some calculators are phones (those very phones), so conclusion (4) holds. Conclusion (1) claims the converse of Statement 2 for calculators→scales, which is not implied; it can be false. Conclusion (3) claims scales ⊆ phones, also not implied.Verification / Alternative check: Example model: phones = {p1}, scales = {p1, s1}, calculators = {p1, s1, c1}. Then (2) and (4) are true, but (1) and (3) are false. Thus only (2) and (4) must follow.Why Other Options Are Wrong: Any option including (1) or (3) assumes a converse or reversal of inclusion not supported by the premises.Common Pitfalls: Believing “All B are C” implies “All C are B”. Always check direction of inclusion.Final Answer: Only (2) and (4)

Question 19

Logical syllogism — determine which conclusions follow beyond doubt.  Statements: 1) Some tables are TV. 2) Some TV are radios.  Conclusions: (1) Some tables are radios. (2) Some radios are tables. (3) All the radios are TV. (4) All the TV are table…

Accepted Answer

Introduction / Context: This question examines intersection reasoning with two “Some …” premises. From two partial overlaps we cannot assert a definite overlap between the first and third sets unless forced by data.Given Data / Assumptions: Some tables ∩ TV ≠ ∅.Some TV ∩ radios ≠ ∅.No universal (All …) statement is given.Concept / Approach: With “Some A are B” statements, the overlapping subsets can be disjoint within B unless specified otherwise. Therefore, we must avoid assuming transitive overlap across two separate “Some …” facts.Step-by-Step Solution: (1) “Some tables are radios” — not compelled. The TV members that are tables may be different individuals from the TV members that are radios. (2) “Some radios are tables” — logically equivalent to (1) by commutation; equally not compelled. (3) “All the radios are TV” — not given; only some TV are radios. (4) “All the TV are tables” — not supported by the partial overlap of tables with TV.Verification / Alternative check: Construct a counterexample: Let TV = {t1, t2}; tables∩TV = {t1}; radios∩TV = {t2}. Then both premises are true, but there is no radio that is a table, and neither universal conclusion holds. Hence none of the four conclusions is necessary.Why Other Options Are Wrong: Options claiming any of (1)–(4) assume more overlap or universality than warranted by the premises.Common Pitfalls: Treating “Some … are …” as if it were transitive. Remember, “some” statements rarely chain deterministically.Final Answer: None of the four

Question 20

Logical syllogism — determine which conclusions follow beyond doubt.  Statements: 1) Some pens are books. 2) Some books are pencils.  Conclusions: (1) Some pens are pencils. (2) Some pencils are pens. (3) All pencils are pens. (4) All books are pens.

Accepted Answer

Introduction / Context: From two independent “Some …” premises, we must avoid inferring intersections that are not guaranteed. This is a common trap in syllogisms.Given Data / Assumptions: Some pens overlap books.Some books overlap pencils.No information connects the same book elements across both statements.Concept / Approach: Separate “Some” overlaps through a middle term (books) do not force an overlap between pens and pencils. Universal conclusions require universal premises, which are absent here.Step-by-Step Solution: (1) “Some pens are pencils” — could be true, but not necessary. The set of books that are pens might differ from the set of books that are pencils. (2) “Some pencils are pens” — equivalent to (1), still not necessary. (3) “All pencils are pens” — far stronger than warranted. (4) “All books are pens” — not implied since only “Some pens are books” is given.Verification / Alternative check: Counterexample: Let books = {b1, b2}; pens∩books = {b1}; pencils∩books = {b2}. Then both premises hold, but pens and pencils do not intersect. Thus none of the conclusions is forced.Why Other Options Are Wrong: Options with (1) or (2) assume an intersection that need not exist. Options with (3) or (4) falsely universalize.Common Pitfalls: Misapplying transitivity to “Some” statements and inferring universals from particulars.Final Answer: None of the four

Syllogism Questions

Questions

Practice smarter