Question 1

Logical syllogism — determine which conclusions follow beyond doubt.  Statements: 1) All the goats are tigers. 2) All the tigers are lions.  Conclusions: (1) All the goats are lions. (2) All the lions are goats. (3) Some lions are goats. (4) Some ti…

Accepted Answer

Introduction / Context: This is a standard three-set syllogism with nested universals. We must derive all conclusions that must be true from the chain goats ⊆ tigers ⊆ lions.Given Data / Assumptions: All goats are tigers.All tigers are lions.Non-emptiness of goats for “Some …” conclusions involving existence.Concept / Approach: Transitivity: if A ⊆ B and B ⊆ C, then A ⊆ C. From “All A are C” and existence of A, we may also infer “Some C are A” and, since A ⊆ B, “Some B are A”.Step-by-Step Solution: (1) All goats are lions — true by transitivity (A ⊆ C). (2) All lions are goats — converse of (1); not supported. (3) Some lions are goats — since goats exist and every goat is a lion, at least some lions are goats. (4) Some tigers are goats — every goat is a tiger; existence of goats guarantees some tigers are goats.Verification / Alternative check: Venn diagram with three nested circles immediately shows (1), (3), and (4) as necessary; (2) is not forced and is generally false unless all lions happen to be goats.Why Other Options Are Wrong: Any option including (2) assumes an unjustified converse.Common Pitfalls: Confusing subset relations with equality and overlooking the existence requirement for “Some …” conclusions.Final Answer: Only (1), (3) and (4)

Question 2

Logical syllogism — determine which conclusions follow beyond doubt.  Statements: 1) All the books are pencils. 2) No pencil is an eraser.  Conclusions: (1) All the pencils are books. (2) Some erasers are books. (3) No book is an eraser. (4) Some bo…

Accepted Answer

Introduction / Context: This syllogism mixes a universal affirmative with a universal negative. We need to deduce what must be true about the relationship between books and erasers.Given Data / Assumptions: All books ⊆ pencils.No pencil is an eraser.Concept / Approach: If a set A is a subset of B, and B is disjoint from C, then A is also disjoint from C. Hence, from “All books are pencils” and “No pencil is an eraser,” we get “No book is an eraser.”Step-by-Step Solution: (3) “No book is an eraser” — follows directly from subset plus disjointness. (1) “All pencils are books” — converse of the first premise; not implied. (2) “Some erasers are books” — contradicts (3); cannot follow. (4) “Some books are erasers” — contradicts (3); cannot follow.Verification / Alternative check: Draw pencils and erasers as disjoint sets; place books entirely within pencils. There is no overlap between books and erasers, proving (3) only.Why Other Options Are Wrong: Options including (1) assume an illegitimate converse; options including (2) or (4) contradict the necessary disjointness.Common Pitfalls: Overlooking that a subset of a set disjoint from C must also be disjoint from C.Final Answer: Only (3)

Question 3

Logical syllogism — determine which conclusions follow beyond doubt.  Statements: 1) All the research scholars are psychologists. 2) Some psychologists are scientists.  Conclusions: (1) All the research scholars are scientists. (2) Some research sch…

Accepted Answer

Introduction / Context: This problem blends a universal inclusion with a particular inclusion. We must identify conclusions that are guaranteed by these relations.Given Data / Assumptions: All research scholars ⊆ psychologists.Some psychologists ⊆ scientists (equivalently, some scientists ⊇ psychologists).Standard existence: research scholars exist for deriving a “Some …” statement involving them.Concept / Approach: Convert the “Some …” statement: “Some psychologists are scientists” implies “Some scientists are psychologists”. Also, from “All research scholars are psychologists,” it is valid (with existence) to state “Some psychologists are research scholars.” However, we cannot say all research scholars are scientists unless the “Some psychologists …” specifically references research scholars, which it does not.Step-by-Step Solution: (3) Some scientists are psychologists — direct conversion of Statement 2; follows. (4) Some psychologists are research scholars — from Statement 1 and existence of research scholars; follows. (1) All research scholars are scientists — not guaranteed; the “some” that are scientists might be different psychologists. (2) Some research scholars are scientists — also not guaranteed without a link connecting the “some” of Statement 2 to research scholars.Verification / Alternative check: Model: psychologists = {a,b,c}; scientists = {a}; research scholars = {b}. Then (1) and (2) fail, but (3) and (4) hold, proving only those two are necessary.Why Other Options Are Wrong: Options including (1) or claiming all four overstate the data. Option (d) omits (4), which is supported under standard exam existence assumptions.Common Pitfalls: Assuming that a “Some …” statement about a superset automatically covers its subset members.Final Answer: Only (3) and (4)

Question 4

Logical syllogism — determine which conclusions follow beyond doubt.  Statements: 1) All the locks are keys. 2) All the keys are bats. 3) Some watches are bats.  Conclusions: (1) Some bats are locks. (2) Some watches are keys. (3) All the keys are l…

Accepted Answer

Introduction / Context: A three-premise syllogism with two universals and one particular checks whether you can keep track of subset chains and avoid invalid converses or unwarranted intersections.Given Data / Assumptions: Locks ⊆ keys.Keys ⊆ bats.Some watches ⊆ bats.Concept / Approach: From transitivity, locks ⊆ bats. “Some watches are bats” does not imply anything about keys unless watches are stated to be keys. Converses like “All keys are locks” are not implied.Step-by-Step Solution: (1) Some bats are locks — since locks ⊆ bats and locks are assumed to exist, there are locks that are bats. Hence true. (2) Some watches are keys — unjustified. While some watches are bats, keys are a subset of bats; the watches could lie outside keys within bats. (3) All the keys are locks — converse of Statement 1; not supported.Verification / Alternative check: Example: bats = {k1, w1}; keys = {k1}; locks = {k1}; watches = {w1}. Then (1) holds, but (2) and (3) fail.Why Other Options Are Wrong: Options including (2) assume an overlap not guaranteed; options including (3) assume a converse.Common Pitfalls: Treating membership in a larger set (bats) as proof of membership in its subset (keys).Final Answer: Only (1)

Question 5

Logical syllogism — determine which conclusions follow beyond doubt.  Statements: 1) Some keys are staplers. 2) Some staplers are stickers. 3) All the stickers are pens.  Conclusions: (1) Some pens are staplers. (2) Some stickers are keys. (3) No st…

Accepted Answer

Introduction / Context: This syllogism combines two particular premises with one universal affirmative about stickers and pens. We need to identify definite conclusions and recognize a classic “either–or” case.Given Data / Assumptions: Some keys ∩ staplers ≠ ∅.Some staplers ∩ stickers ≠ ∅.All stickers ⊆ pens.Concept / Approach: From “Some staplers are stickers” and “All stickers are pens,” the particular set of staplers that are stickers is also a subset of pens. Therefore, there exists at least one element that is both a pen and a stapler. Additionally, “Some keys are staplers” immediately yields “Some staplers are keys.” However, with respect to the relation between stickers and keys, the data permit either overlap or separation; thus an exclusive uncertainty arises: either some stickers are keys, or no sticker is a key, but we cannot fix which.Step-by-Step Solution: (4) “Some staplers are keys” — directly from Statement 1 by commutation; this must follow. (1) “Some pens are staplers” — since some staplers are stickers and all stickers are pens, those staplers are pens; hence some pens are staplers. (2) “Some stickers are keys” — not forced; could be true if the overlapping staplers are among the key-staplers. (3) “No sticker is key” — also not forced; could be true if the sticker-staplers are not among the key-staplers.Verification / Alternative check: Construct two models: Model A with an element that is key+stapler+sticker (then (2) true, (3) false); Model B where the staple-sticker element is different from the key-stapler element (then (3) true, (2) false). In both models, (1) and (4) are always true. Hence the definite outcome is (1) and (4), and either (2) or (3).Why Other Options Are Wrong: Options that pick only (1) and (2) or only (2) and (4) ignore the alternative possibility. The only option capturing the certainty plus the exclusive uncertainty is the “Only (1) and (4) and either (2) or (3)” choice.Common Pitfalls: Assuming transitivity of “Some …” overlaps through a middle term without checking whether the same elements are involved.Final Answer: Only (1) and (4) and either (2) or (3)

Question 6

Syllogism — Validate Conclusions from Given Statements Statements: • Some questions are answers. • Some answers are writers. • All writers are poets. Conclusions: (1) Some writers are answers. (2) Some poets are questions. (3) All the questions are …

Accepted Answer

Introduction / Context:Syllogism questions test whether a conclusion must be true based strictly on the given statements. We cannot add outside facts, and we must avoid assuming overlaps unless they are logically compelled by the premises.Given Data / Assumptions:Some questions are answers.Some answers are writers.All writers are poets.Conclusions to evaluate: (1) Some writers are answers. (2) Some poets are questions. (3) All the questions are poets. (4) Some poets are answers.Concept / Approach:Use classic set/venn reasoning. “All X are Y” means X is a subset of Y. “Some X are Y” indicates a definite non-empty intersection. Only conclusions that must hold for all diagrams consistent with the premises are valid.Step-by-Step Solution:From “Some answers are writers,” we can also say “Some writers are answers.” This validates conclusion (1).All writers are poets. The “some answers that are writers” are therefore also poets. Hence there exist entities that are both poets and answers, validating conclusion (4).For (2): Some poets are questions? We know some questions are answers, and some answers are writers, but the two “some” groups need not be the same individuals. An overlap between “questions” and “writers” (and thus poets) is not guaranteed. Hence (2) does not necessarily follow.For (3): All questions are poets is far stronger than warranted. We only know “some questions are answers,” not that all questions connect to writers/poets. So (3) is invalid.Verification / Alternative check:Draw Venn sets for Questions (Q), Answers (A), Writers (W), and Poets (P) with W entirely inside P. Place one element in A∩W (proves (1) and (4)). Place a separate element in Q∩A that is not in W. This diagram satisfies premises while falsifying (2) and (3).Why Other Options Are Wrong:Only (1) and (2): (2) is not forced.Only (1) and (3): (3) is too strong.Only (2) and (4): (2) fails.None of the above: Incorrect because (1) and (4) do follow.Common Pitfalls:Assuming that “some … are …” chains guarantee overlap across multiple “some” statements. They do not. Only a single “all” statement permits a definite subset relation.Final Answer:Only (1) and (4)

Question 7

Syllogism — Envelopes, Gums, Seals, and Adhesives Statements: • Some envelopes are gums. • Some gums are seals. • Some seals are adhesives. Conclusions: (1) Some envelopes are seals. (2) Some gums are adhesives. (3) Some adhesives are seals. (4) Som…

Accepted Answer

Introduction / Context:This problem asks which conclusions necessarily follow given three “some” statements across four categories. With “some,” overlaps are possible but not guaranteed unless directly stated or logically converted.Given Data / Assumptions:E = envelopes, G = gums, S = seals, A = adhesives.Premises: Some E are G; Some G are S; Some S are A.Conclusions to test: (1) Some E are S; (2) Some G are A; (3) Some A are S; (4) Some A are G.Concept / Approach:Only conclusions that must be true in every diagram consistent with the premises are valid. From “Some S are A,” we can apply conversion for “some”: “Some A are S.” Other claimed overlaps require the same individuals to be shared across different “some” statements, which is not forced.Step-by-Step Solution:(3) Some adhesives are seals: Directly from “Some seals are adhesives,” convert to “Some adhesives are seals.” Valid.(1) Some envelopes are seals: We have E∩G ≠ ∅ and G∩S ≠ ∅, but the overlapping members of G in each statement need not be the same element. So E∩S may be empty. Not guaranteed.(2) Some gums are adhesives: Similarly, S∩A ≠ ∅ and G∩S ≠ ∅ do not force G∩A ≠ ∅. Not guaranteed.(4) Some adhesives are gums: Again, nothing compels A to overlap with G.Verification / Alternative check:Construct a counterexample: place one item in E∩G; a different item in G∩S; and another in S∩A. All premises hold while E∩S, G∩A, and E∩A remain empty. Thus only (3) must follow.Why Other Options Are Wrong:(1), (2), (4) each assumes compelled overlap among different “some” statements. This is not logically necessary.Only (3) is the unique must-follow conclusion.Common Pitfalls:Chaining “some” statements as if they behave like “all.” Remember: “some” + “some” does not compel intersection beyond what is explicitly stated.Final Answer:Only (3)

Question 8

Syllogism — Papers, Bags, Books, and Purses Statements: • All the papers are books. • All the bags are books. • Some purses are bags. Conclusions: (1) Some papers are bags. (2) Some books are papers. (3) Some books are purses.

Accepted Answer

Introduction / Context:Here we mix two universal statements with one particular statement. The challenge is to see which particular conclusions are compelled by subset relations and which are not forced due to possible non-overlap.Given Data / Assumptions:Papers (P) ⊆ Books (B).Bags (G) ⊆ Books (B).Some Purses (R) are Bags (G): R∩G ≠ ∅.Conclusions: (1) Some papers are bags. (2) Some books are papers. (3) Some books are purses.Concept / Approach:“All X are Y” creates subset relations. From a subset relation we can assert “Some Y are X” when existence is implicit in such exam settings. For intersection claims across two subsets of the same superset, we cannot assume overlap unless given or compelled.Step-by-Step Solution:(2) Some books are papers: Since all papers are books and papers exist, at least some books are indeed papers. Valid.(3) Some books are purses: We are told some purses are bags, and all bags are books. Hence those purses are books too. So some books are purses. Valid.(1) Some papers are bags: Although both papers and bags are subsets of books, they can be disjoint within B. Nothing forces an overlap. Not valid.Verification / Alternative check:Draw B as a large set containing two non-overlapping regions P and G. Place some elements in R∩G. Then (2) and (3) are satisfied; (1) remains false. All premises still hold, proving (1) does not follow.Why Other Options Are Wrong:Only (1): Fails because (1) is not necessary.Only (1) and (2) / Only (1) and (3): Both include (1), which does not follow.None of the above: Incorrect because (2) and (3) do follow.Common Pitfalls:Assuming that two subsets of the same superset must intersect. They need not. Only explicit intersection statements compel that conclusion.Final Answer:Only (2) and (3)

Question 9

Syllogism — Rats, Cats, Dogs, and Cows Statements: • Some rats are cats. • Some cats are dogs. • No dog is a cow. Conclusions: (1) No cow is a cat. (2) No dog is a rat. (3) Some cats are rats.

Accepted Answer

Introduction / Context:This syllogism mixes “some” and a categorical negative (“No dog is a cow”). The task is to check which conclusions must follow without overreaching beyond what is guaranteed by the premises.Given Data / Assumptions:Some Rats (R) are Cats (C): R∩C ≠ ∅.Some Cats (C) are Dogs (D): C∩D ≠ ∅.No Dog (D) is a Cow (W): D∩W = ∅.Conclusions to test: (1) No Cow is a Cat. (2) No Dog is a Rat. (3) Some Cats are Rats.Concept / Approach:Only conclusions that hold for all possible diagrams are valid. A single counterexample diagram that satisfies premises but falsifies a conclusion is enough to reject it.Step-by-Step Solution:(3) Some cats are rats: Direct restatement of the first premise (“Some rats are cats”). Valid.(1) No cow is a cat: We only know that no dog is a cow and that some cats overlap with dogs. This says nothing about the remaining cats that are not dogs; they could overlap with cows or not. Hence (1) is not compelled.(2) No dog is a rat: From the premises, it is even possible that the “some cats that are dogs” are the same as the “some rats that are cats,” giving a dog that is also a rat. Therefore the universal negative (2) is not guaranteed.Verification / Alternative check:Construct a diagram where one element lies in R∩C∩D (allowed by the first two premises) and D and W remain disjoint. All premises hold; (2) becomes false, confirming it does not follow. Similarly, allow some C to overlap W (not prohibited). This falsifies (1) while satisfying premises.Why Other Options Are Wrong:Any option containing (1) or (2) overstates what is given.Only (3) is supported with certainty.Common Pitfalls:Assuming that if some cats are dogs and no dog is a cow, then no cat is a cow. That generalization is invalid because “some cats” does not describe all cats.Final Answer:Only (3)

Question 10

Syllogism — Books, Papers, Journals, and Calendars Statements: • All the books are papers. • Some papers are journals. • Some journals are calendars. Conclusions: (1) Some journals are books. (2) Some calendars are papers. (3) Some books are journal…

Accepted Answer

Introduction / Context:This item explores how multiple subset and “some” statements interact. The trap is to assume overlaps that are not guaranteed by the premises, especially when two separate “some” statements could refer to different subgroups.Given Data / Assumptions:Bk ⊆ Pa (All books are papers).Some Pa are Jo (some papers are journals).Some Jo are Ca (some journals are calendars).Conclusions: (1) Some journals are books. (2) Some calendars are papers. (3) Some books are journals. (4) Some books are calendars.Concept / Approach:We must test whether a conclusion must be true in every model satisfying the premises. If a counterexample exists, that conclusion does not follow. Separate “some” statements may refer to disjoint elements unless a link forces the same individuals to be shared.Step-by-Step Solution:Arrange Pa as a large set. Place Bk as a subset of Pa, disjoint from Jo if desired (allowed by premises).Add one element in Pa∩Jo to satisfy “some papers are journals.” This element need not be a book.Add a different element in Jo∩Ca to satisfy “some journals are calendars.” This need not be within Pa, because nothing states that all journals are papers.This model satisfies all premises while making Bk∩Jo empty and Pa∩Ca empty. Therefore (1), (2), (3), and (4) all fail.Verification / Alternative check:Because nothing requires the “some papers that are journals” to be the same individuals as the “some journals that are calendars,” overlaps with books or calendars are not forced. Hence none of the conclusions must follow.Why Other Options Are Wrong:(1), (3), (4) rely on an unforced overlap between Books and Journals (or Calendars).(2) assumes Journals that are Calendars must also be Papers; the premises do not require that.Common Pitfalls:Chaining separate “some” facts as if they refer to the same elements. Without an “all” or explicit identity, such chaining is invalid.Final Answer:None of the four

Question 11

Syllogism — Bottles, Boxes, Bags, and Trays Statements: • All the bottles are boxes. • All the boxes are bags. • Some bags are trays. Conclusions: (1) Some bottles are trays. (2) Some trays are boxes. (3) All the bottles are bags. (4) Some trays are…

Accepted Answer

Introduction / Context:This is a classic categorical syllogism with two universals and one particular. The goal is to see which conclusions are compelled by subset relations and which are not forced by the existence of “some” in a different subset.Given Data / Assumptions:Bottles (Bo) ⊆ Boxes (Bx).Boxes (Bx) ⊆ Bags (Bg).Some Bags (Bg) are Trays (Tr): Bg∩Tr ≠ ∅.Conclusions to check: (1) Some Bo are Tr. (2) Some Tr are Bx. (3) All Bo are Bg. (4) Some Tr are Bg.Concept / Approach:Transitivity of subsets: from Bo ⊆ Bx and Bx ⊆ Bg, we get Bo ⊆ Bg. Particular intersections can be converted: if some Bg are Tr, then some Tr are Bg (but not necessarily boxes or bottles).Step-by-Step Solution:(3) All the bottles are bags: By transitivity, Bo ⊆ Bx ⊆ Bg ⇒ Bo ⊆ Bg. Valid.(4) Some trays are bags: Given “Some bags are trays,” by conversion of “some,” “Some trays are bags.” Valid.(1) Some bottles are trays: Not forced. The trays that overlap bags may all lie outside the bottles subset of bags.(2) Some trays are boxes: Similarly, trays need not overlap boxes; they only must overlap bags.Verification / Alternative check:Construct a diagram with Bo inside Bx inside Bg. Place a few Tr elements intersecting Bg but outside Bx. Premises hold; (1) and (2) are false, confirming only (3) and (4) must follow.Why Other Options Are Wrong:Options including (1) or (2) assume unnecessary overlaps with bottles/boxes.Only the pair (3) and (4) is guaranteed.Common Pitfalls:Assuming that because trays intersect bags, they must also intersect every subset within bags. That is not logically required.Final Answer:Only (3) and (4)

Question 12

Syllogism — Cars, Jeeps, Boxes, and Pens Statements: • Some cars are jeeps. • All the boxes are jeeps. • All the pens are cars. Conclusions: (1) Some cars are boxes. (2) No pen is a jeep. (3) Some boxes are cars.

Accepted Answer

Introduction / Context:This question mixes “some” with two universal statements. The danger is to infer overlaps between Cars and Boxes or Pens and Jeeps without explicit compulsion.Given Data / Assumptions:Some Cars (C) are Jeeps (J): C∩J ≠ ∅.All Boxes (B) are Jeeps (J): B ⊆ J.All Pens (P) are Cars (C): P ⊆ C.Conclusions: (1) Some C are B. (2) No P is J. (3) Some B are C.Concept / Approach:To prove “some X are Y,” we need a compelled intersection. To prove “No X is Y,” we need a compelled disjointness. Here, neither is forced between Cars and Boxes nor between Pens and Jeeps.Step-by-Step Solution:Because B ⊆ J and only “some” Cars are Jeeps, Boxes could lie wholly in the Jeep region disjoint from the Car portion. Hence neither “Some cars are boxes” nor “Some boxes are cars” is compelled.All Pens are Cars, yet some Cars are Jeeps. Pens might be in the non-Jeep area of Cars or could also overlap Jeeps; premises do not fix this. Therefore “No pen is a jeep” is not guaranteed.Verification / Alternative check:Draw J as a set. Place B entirely inside J but outside C. Place C partially overlapping J (to satisfy “some C are J”) and put P inside the C area that is outside J. All premises hold; (1), (2), and (3) are false simultaneously, proving that none of the three conclusions must follow.Why Other Options Are Wrong:Pairs like (1)&(2) or (1)&(3) assume overlaps not forced by the statements.(2)&(3) also fails because (2) is not compelled and (3) is not compelled either.Common Pitfalls:Assuming that two sets contained in a third must overlap each other. Subsets of a common superset can still be disjoint.Final Answer:None of three

Question 13

Syllogism — Evaluate the conclusions from the following statements: Statements: • All pins are rods. • Some rods are chains. • All chains are hammers. Conclusions: I. Some pins are hammers. II. Some hammers are rods. III. No pin is a hammer. Choose …

Accepted Answer

Introduction / Context:This question tests classical categorical syllogism. We must determine which conclusions necessarily follow from the given premises, not what could be true. The terms are “pins,” “rods,” “chains,” and “hammers.”Given Data / Assumptions:All pins are rods.Some rods are chains.All chains are hammers.Concept / Approach:Translate the statements into set relations. “All X are Y” means X ⊆ Y. “Some X are Y” indicates a non-empty intersection. Only conclusions that are true in every model consistent with the premises can be marked as “follows.”Step-by-Step Solution:1) From “Some rods are chains” and “All chains are hammers,” we get: some rods are hammers.2) From “All pins are rods,” pins are a subset of rods, but there is no claim that any pins are among those rods that are chains.3) Therefore, it is possible that no pin belongs to chains (and thus to hammers). Hence “Some pins are hammers” is not necessary.4) Likewise, “No pin is a hammer” is not necessary either because pins could overlap with the chain portion of rods in some models.5) However, “Some hammers are rods” is guaranteed because those rods that are chains are also hammers.Verification / Alternative check:Construct two models: (a) pins entirely outside chains (I false, II true, III true); (b) pins overlapping chains (I true, II true, III false). Only conclusion II remains true across all models.Why Other Options Are Wrong:I (“Some pins are hammers”): possible but not necessary.III (“No pin is a hammer”): possible but not necessary.“Either I or III” with II: the “either-or” part is not guaranteed; only II is certain.Common Pitfalls:Confusing “can be” with “must be,” and assuming subsets necessarily intersect without explicit information.Final Answer:Only II follows.

Question 14

Syllogism — Determine the valid conclusions from the following: Statements: • Some books are papers. • Some papers are desks. • All desks are chairs. Conclusions: I. Some books are desks. II. Some papers are chairs. III. Some books are chairs.

Accepted Answer

Introduction / Context:The task is to test necessity of conclusions from three premises mixing “some” (particular) and “all” (universal) statements.Given Data / Assumptions:Some books are papers.Some papers are desks.All desks are chairs.Concept / Approach:Only conclusions that hold in every permissible diagram (Venn model) are valid. Particular statements certify existence; universal statements give subset relations.Step-by-Step Solution:1) “Some papers are desks” and “All desks are chairs” imply “Some papers are chairs.” So II is necessary.2) I (“Some books are desks”) would require overlap among the specific “books∩papers” and “papers∩desks.” The two “some” portions might be disjoint; thus I is not necessary.3) III (“Some books are chairs”) would require the “books∩papers” part to lie within desks (then chairs). That is not forced; hence III is not necessary.Verification / Alternative check:Create a model where the “books∩papers” area is separate from the “papers∩desks” area. II remains true, I and III fail—confirming only II follows.Why Other Options Are Wrong:“None follows”: II clearly follows.“Only I” / “Only III”: unsupported.“All follow”: overstates the logical force.Common Pitfalls:Assuming all “some” parts coincide; they need not.Final Answer:Only II follows.

Question 15

Syllogism — Judge which conclusions must follow: Statements: • Some pots are buckets. • All buckets are tubs. • All tubs are drums. Conclusions: I. Some drums are pots. II. All tubs are buckets. III. Some drums are buckets.

Accepted Answer

Introduction / Context:This is a chain of subset relations with one existential (“Some pots are buckets”). We must decide which conclusions necessarily hold.Given Data / Assumptions:Some pots are buckets (pots∩buckets ≠ ∅).All buckets are tubs (buckets ⊆ tubs).All tubs are drums (tubs ⊆ drums).Concept / Approach:Use transitivity of subsets and track the existential element guaranteed by “Some pots are buckets.”Step-by-Step Solution:1) From “Some pots are buckets” and the chain buckets ⊆ tubs ⊆ drums, those specific pots are also drums. Hence “Some drums are pots” is necessary (Conclusion I true).2) “All tubs are buckets” is the converse of the given “All buckets are tubs.” Converse need not hold; Conclusion II is false.3) Because buckets ⊆ drums and at least one bucket exists (from “Some pots are buckets”), there exists at least one object that is both a drum and a bucket. Hence “Some drums are buckets” is necessary (Conclusion III true).Verification / Alternative check:Draw a chain Pots → Buckets → Tubs → Drums with a marked example element in pots∩buckets. The element traces into drums, confirming I and III.Why Other Options Are Wrong:“Only I and II”/“Only II and III”/“All”: each wrongly includes II.Common Pitfalls:Assuming converses of universal statements and overlooking the existential support provided by “Some…”Final Answer:Only I and III follows.

Question 16

Syllogism — Decide which conclusion(s) necessarily follow: Statements: • All pins are bags. • All chalks are bags. • All needles are bags. Conclusions: I. Some needles are pins. II. Some chalks are needles. III. No needle is a pin.

Accepted Answer

Introduction / Context:All three categories (pins, chalks, needles) are placed inside a super-set “bags.” The prompt tests recognition of an “either–or” situation with no forced overlap.Given Data / Assumptions:All pins are bags.All chalks are bags.All needles are bags.Concept / Approach:When multiple classes are only related to a common superset and no intersections are specified, two sub-classes may or may not overlap. A conclusion “Some X are Y” is not necessary; “No X is Y” is not necessary either. However, logically, either overlap exists or it does not—making an exclusive “either–or” statement valid in many test conventions.Step-by-Step Solution:1) From the premises, we cannot force “needles∩pins ≠ ∅” (so I is not necessary).2) Nor can we force “needles∩pins = ∅” (so III is not necessary individually).3) But exactly one of I or III must hold in any model: either there is some overlap or there is none. Hence the standard syllogism key accepts “Either I or III follows.”4) II (“Some chalks are needles”) has no support.Verification / Alternative check:Construct two diagrams: (a) needles overlapping pins, (b) needles disjoint from pins. Premises are satisfied in both, but I and III cannot be simultaneously true—thereby validating the exclusive either–or form.Why Other Options Are Wrong:I only / III only: neither is forced alone.Including II: unjustified; chalks and needles need not intersect.Common Pitfalls:Assuming overlap just because classes share a common superset; equating “possibility” with “necessity.”Final Answer:Only either I or III follows.

Question 17

Syllogism — Choose the conclusion(s) that must follow: Statements: • Some buses are trucks. • Some trucks are boats. • No boat is a jeep. Conclusions: I. Some jeeps are buses. II. Some boats are buses. III. Some jeeps are trucks.

Accepted Answer

Introduction / Context:This item checks whether you can avoid over-linking two independent “some” statements and handle a negative statement correctly.Given Data / Assumptions:Some buses are trucks.Some trucks are boats.No boat is a jeep.Concept / Approach:Two separate “some” premises do not force their intersections to meet. Also, “No boat is a jeep” only restricts boats relative to jeeps; it says nothing about other trucks or buses.Step-by-Step Solution:1) I (“Some jeeps are buses”): no link establishes jeeps with buses—cannot be concluded.2) II (“Some boats are buses”): the “some buses are trucks” set may be different from the “some trucks are boats” set; no necessity of overlap—cannot be concluded.3) III (“Some jeeps are trucks”): not supported; the negative premise only says boats are not jeeps.Verification / Alternative check:Draw a model where the truck portion that is a bus is disjoint from the truck portion that is a boat. All premises hold; none of I–III follow.Why Other Options Are Wrong:Each asserts a relationship not compelled by the premises.Common Pitfalls:Assuming separate “some” facts refer to the same elements; misusing the negative statement to claim more than it states.Final Answer:None follows.

Question 18

Syllogism — Determine which conclusions necessarily follow: Statements: • All windows are doors. • All buildings are doors. • All doors are boats. Conclusions: I. All windows are boats. II. All buildings are boats. III. Some boats are doors.

Accepted Answer

Introduction / Context:A classic transitivity problem where two classes (windows, buildings) both sit inside “doors,” and doors sit inside “boats.” We must test three conclusions.Given Data / Assumptions:Windows ⊆ Doors.Buildings ⊆ Doors.Doors ⊆ Boats.Concept / Approach:For universal statements, subset transitivity applies: if X ⊆ Y and Y ⊆ Z, then X ⊆ Z.Step-by-Step Solution:1) I: Windows ⊆ Doors and Doors ⊆ Boats ⇒ Windows ⊆ Boats. True.2) II: Buildings ⊆ Doors and Doors ⊆ Boats ⇒ Buildings ⊆ Boats. True.3) III: Since every door is a boat, provided doors exist, at least “some boats are doors.” In standard exam convention with universal chains like these, this is accepted as true.Verification / Alternative check:Diagram three nested sets: Windows and Buildings inside Doors, Doors inside Boats; the results are immediate.Why Other Options Are Wrong:Any option omitting one of I/II/III underestimates the force of the universal chain.Common Pitfalls:Overlooking that if Doors are Boats, then every existing Door is a Boat, which licenses the particular statement “Some boats are doors” under typical test assumptions.Final Answer:All follow.

Question 19

Syllogism — From the statements below, identify the correct conclusion(s): Statements: I. Some keys are locks; some locks are numbers. II. All numbers are letters; all letters are words. Conclusions: I. Some words are numbers. II. Some locks are let…

Accepted Answer

Introduction / Context:This problem combines particular and universal premises across five terms. We must check two candidate conclusions.Given Data / Assumptions:Some keys are locks; some locks are numbers.All numbers are letters; all letters are words.Concept / Approach:Use subset chaining for universals and existence from particulars. If Numbers ⊆ Letters ⊆ Words, then every Number is a Word.Step-by-Step Solution:1) Because all Numbers are Words, if any Numbers exist, then some Words are Numbers.2) Existence of Numbers is guaranteed by “some locks are numbers.” Hence Conclusion I is necessary.3) Those Locks that are Numbers are also Letters (Numbers ⊆ Letters). Therefore “Some locks are letters” is necessary (Conclusion II).Verification / Alternative check:Mark an element that is both Lock and Number; propagate along the universal chain to confirm it is a Letter and therefore a Word.Why Other Options Are Wrong:Any option omitting I or II ignores either the existential support or the universal chain.Common Pitfalls:Failing to leverage the guaranteed existence from a “some” premise before applying universal inclusions.Final Answer:Conclusions I and II follow.

Question 20

Syllogism — Assess the inference from the given statements: Statements: • All professors are learned. • All learned people are gentle. Inference to test: All professors are gentle persons.

Accepted Answer

Introduction / Context:This is a straightforward two-step subset chain. We must decide whether the stated inference is a necessary consequence of the premises.Given Data / Assumptions:All professors are learned.All learned people are gentle.Concept / Approach:If A ⊆ B and B ⊆ C, then A ⊆ C. Here: Professors ⊆ Learned ⊆ Gentle.Step-by-Step Solution:1) From “All professors are learned,” every professor is in the set Learned.2) From “All learned are gentle,” every member of Learned is in Gentle.3) Therefore every Professor is in Gentle, which is exactly the tested inference.Verification / Alternative check:Any Venn diagram with concentric containment Learned inside Gentle and Professors inside Learned validates the inference.Why Other Options Are Wrong:“False”: contradicts the transitive inclusion.“Probably true/false” or “irrelevant”: this is a deductive certainty, not probabilistic.Common Pitfalls:Overcomplicating a direct chain; forgetting that two universal inclusions compose into a third.Final Answer:The inference is true.

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