Question 1

Syllogism — combine two “some” statements with a shared middle class  Statements: • Some houses are offices. • Some offices are schools. Conclusions to evaluate: I. Some schools are houses. II. Some offices are houses. III. No house is school. IV. S…

Accepted Answer

Introduction / Context:This exercise assesses how to read two independent some-type premises that meet at a middle term. The first links houses and offices; the second links offices and schools. The question is which conclusions are guaranteed without overreaching.Given Data / Assumptions:H = houses, O = offices, S = schools.Some H are O.Some O are S.Concept / Approach:From some A are B you may always infer the symmetric some B are A, because it asserts the existence of elements in the intersection. Therefore, from some H are O, we also have some O are H (conclusion II). Likewise, from some O are S we get some S are O (conclusion IV). However, the two premises do not force a link between H and S, so neither an overlap nor a disjointness between H and S is necessary.Step-by-Step Solution:Conclusion II: Some offices are houses is the symmetric form of the first premise. Hence it follows.Conclusion IV: Some schools are offices is the symmetric form of the second premise. Hence it follows.Conclusion I: Some schools are houses would require the two intersections to share common elements. That need not be the case; it does not follow.Conclusion III: No house is school is an unwarranted universal negative. The premises do not assert such exclusion; it does not follow.Verification / Alternative check:Model: Let O contain two disjoint subsets, O1 overlapping H and O2 overlapping S, with O1 ∩ O2 = ∅. Then some H are O and some O are S both hold. II and IV are true (by symmetry), while I and III are false, confirming the result.Why Other Options Are Wrong:Only I and II / Only I and IV: each includes I which is not compelled.Only II follows: omits IV, which is equally guaranteed.None follows: incorrect because II and IV are necessary.Common Pitfalls:Illicitly chaining some-premises to force an H–S overlap; assuming a universal negative without evidence.Final Answer:Only II and IV follow

Question 2

Syllogism with two independent properties — horns and lights on taxis  Statements: • Some taxis have horns. • Some taxis have lights. Conclusions to evaluate: I. Every taxi has either a horn or a light. II. Some taxis have neither a light nor a horn…

Accepted Answer

Introduction / Context:This question examines reasoning about existence with two independent features on the same class (taxis). The premises say at least one taxi has a horn and at least one taxi has a light, but they do not specify whether any taxi has both or whether some have neither.Given Data / Assumptions:T = set of taxis, H = taxis with horns, L = taxis with lights.H is nonempty; L is nonempty.No other restrictions on H and L are given.Concept / Approach:From the premises alone, two mutually exclusive possibilities remain about the intersection H ∩ L: either it is nonempty (at least one taxi has both features) or it is empty (no taxi has both). Logic guarantees that exactly one of these two possibilities must be true. However, nothing forces every taxi to have at least one feature or guarantees that some taxis have neither.Step-by-Step Solution:Conclusion I is too strong; the premises do not cover all taxis. So I does not follow.Conclusion II is not forced; there may be zero taxis lacking both features.Conclusion III states that the intersection H ∩ L is nonempty. This may be true, but it is not compelled.Conclusion IV claims the intersection is empty. This also may be true, but it is not compelled.Since exactly one of III or IV must be true in any model (the intersection is either empty or not), the disjunctive conclusion “Either III or IV follows” is necessarily correct, even though we cannot say which one individually is true from the premises.Verification / Alternative check:Model A: One taxi is both horned and lighted (intersection nonempty) shows III true and IV false. Model B: Different taxis separately carry horn and light with no overlap shows III false and IV true. In both models, I and II remain unforced.Why Other Options Are Wrong:Pairs like I and II / II and III / II and IV: each asserts non-necessary claims.All follow: clearly false since I and II are not guaranteed.Common Pitfalls:Assuming that some in both features implies there is a taxi with both; assuming every taxi must have at least one highlighted feature.Final Answer:Either III or IV follows

Question 3

Syllogism — chaining universal inclusions across three classes  Statements: • All fruits are vegetables. • All pens are vegetables. • All vegetables are rains. Conclusions to evaluate: I. All fruits are rains. II. All pens are rains. III. Some rains…

Accepted Answer

Introduction / Context:This syllogism features a clean chain of subset relations. The task is to apply transitivity and, where appropriate, infer safe existential statements from universal ones when nonemptiness is ensured by earlier premises.Given Data / Assumptions:F = fruits, V = vegetables, R = rains, Pn = pens.F ⊆ V and Pn ⊆ V.V ⊆ R.F is nonempty (as a named real-world category), hence V is nonempty as well.Concept / Approach:Transitivity of inclusion gives immediate universal conclusions about fruits and pens relative to rains. In addition, since V is nonempty and V ⊆ R, there exists at least one element that is both rain and vegetable, which justifies an existential conclusion.Step-by-Step Solution:Conclusion I: From F ⊆ V and V ⊆ R, transitivity yields F ⊆ R. Therefore all fruits are rains. I follows.Conclusion II: From Pn ⊆ V and V ⊆ R, we get Pn ⊆ R. Therefore all pens are rains. II follows.Conclusion III: Because V is nonempty and V ⊆ R, pick any v in V; then v is a vegetable and also rain. Hence some rains are vegetables. III follows.Verification / Alternative check:Diagrammatically, put F and Pn inside V and then place V fully inside R. This immediately shows I and II. The nonemptiness of V guarantees at least one overlap element for III.Why Other Options Are Wrong:Options B, C, and D exclude one of the valid conclusions.Option A claims none follows, which contradicts clear transitivity results.Common Pitfalls:Forgetting to apply transitivity; hesitation about III due to universal premises—remember that the existence of fruits implies vegetables exist, enabling the existential claim.Final Answer:All follow

Question 4

Syllogism — mixing “some”, “no”, and a universal inclusion  Statements: • Some towels are brushes. • No brush is soap. • All soaps are rats. Conclusions to evaluate: I. Some rats are brushes. II. No rat is brush. III. Some towels are soaps.

Accepted Answer

Introduction / Context:This question blends an existential overlap with a universal negative and a universal inclusion. The aim is to see whether any necessary relationship can be forced between brushes and rats or between towels and soaps.Given Data / Assumptions:T = towels, B = brushes, S = soaps, R = rats.Some T are B.No B is S.All S are R.Concept / Approach:No B is S tells us brushes and soaps are disjoint. All S are R only places soaps inside rats; it does not say all rats are soaps. Therefore, brushes might still be rats (if there exist rats outside S) or might not be; the premises are silent. Also, nothing connects towels to soaps except through brushes, and brushes are explicitly not soaps.Step-by-Step Solution:Conclusion I (Some rats are brushes) is not forced. It would require B ∩ R to be nonempty. Since R may contain elements beyond S, the premises neither ensure nor forbid such an overlap.Conclusion II (No rat is brush) is also not compelled. The premises only deny B ∩ S, not B ∩ R. If R includes non-soap rats, brushes could be among those.Conclusion III (Some towels are soaps) fails because the only link from towels to soaps would go through brushes (some towels are brushes), yet no brush is a soap.Verification / Alternative check:Model A: Let S ⊆ R and R also contain r1 not in S. Let B = {r1} and some T include r1. Then I is true and II is false. Model B: Let B disjoint from R entirely, and T overlap B. Then II is true and I false. Since different valid models yield different truth values, none of I–III is necessary.Why Other Options Are Wrong:Only either I or II follows: although exactly one of them will be true in any single model, the premises do not force which one; therefore the disjunction is not a necessary consequence here.Only II follows / Only I and III follow: each asserts claims that can fail under valid models.None of these: option A already captures that no listed conclusion must hold.Common Pitfalls:Assuming all rats are soaps (converse of all S are R); trying to pass through the brush set to reach soaps despite the explicit exclusion.Final Answer:None follows

Question 5

Syllogism – Determine which conclusions logically follow Statements: • Some pictures are frames. • Some frames are idols. • All idols are curtains. Conclusions to test: I. Some curtains are pictures. II. Some curtains are frames. III. Some idols are…

Accepted Answer

Introduction / Context:This problem tests standard categorical logic (syllogism) using set relations such as “all,” “some,” and “no.” You must decide which conclusions are guaranteed by the three given statements without injecting extra assumptions. Visualizing with simple Venn-style overlaps or set inclusion helps prevent common traps.Given Data / Assumptions:Some pictures are frames (P ∩ F is non-empty).Some frames are idols (F ∩ I is non-empty).All idols are curtains (I ⊆ C).“Some” asserts existence; “all” indicates subset containment.Concept / Approach:Convert each English statement into set relations and propagate inclusion: when “all idols are curtains,” every idol element also belongs to curtains. When “some frames are idols,” the very same elements are frames and idols simultaneously; by inclusion, they are also curtains. Be careful: two different “some” statements need not refer to the same specific elements.Step-by-Step Solution:Check II: “Some curtains are frames.” From “some frames are idols” and “all idols are curtains,” the frame elements that are idols are automatically curtains. Therefore at least one element is both curtain and frame. II follows.Check III: “Some idols are frames.” This is the direct converse of “some frames are idols.” With “some,” conversion is valid: if some F are I, then some I are F. III follows.Check I: “Some curtains are pictures.” We know some frames are idols and thus curtains, and (separately) some pictures are frames. But these two “some” groups need not be the same frames. There is no forced overlap between pictures and idols. Hence I is not guaranteed.Verification / Alternative check:Create a model where one subset of frames overlaps idols (becoming curtains) and a different subset of frames overlaps pictures, with those subsets disjoint. All premises hold while I fails, confirming that only II and III must follow.Why Other Options Are Wrong:I and II: includes I, which is not compelled.I and III: again relies on unforced overlap between pictures and curtains.All follow: overly strong; I fails in valid models.None of these: incorrect because II and III do follow.Common Pitfalls:Assuming two separate “some” statements point to the same elements; assuming transitivity for “some” across different pairs without evidence.Final Answer:Only II and III follow

Question 6

Syllogism – Evaluate conclusions from partial overlaps Statements: • Some hills are rivers. • Some rivers are deserts. • All deserts are roads. Conclusions to test: I. Some roads are rivers. II. Some roads are hills. III. Some deserts are hills.

Accepted Answer

Introduction / Context:Syllogism questions often combine “some” (existential overlap) with an “all” (subset) statement. The challenge is to propagate definite implications while resisting the urge to assume extra intersections that are not forced by the data.Given Data / Assumptions:Some hills are rivers (H ∩ R ≠ ∅).Some rivers are deserts (R ∩ D ≠ ∅).All deserts are roads (D ⊆ Rd).Conclusions about “some” require a guaranteed witness element from the premises.Concept / Approach:Use set inclusion and overlaps: from R ∩ D and D ⊆ Rd, we can chain membership to show at least one element is both a river and a road. But to connect hills with deserts or roads, we would need a shared element proven to be in both overlaps, which is not provided.Step-by-Step Solution:Conclusion I: From “some rivers are deserts” and “all deserts are roads,” the specific elements that are both river and desert are also roads. Therefore, “some roads are rivers” is guaranteed. I follows.Conclusion II: To claim “some roads are hills,” we would need an element shown to be both a hill and a road. We have H ∩ R ≠ ∅ and R ∩ D ≠ ∅, but the hill–river elements could be different from the river–desert elements. No necessary overlap among hills and deserts is given. II does not follow.Conclusion III: “Some deserts are hills” would require D ∩ H ≠ ∅. No statement links deserts directly with hills; it is not guaranteed. III does not follow.Verification / Alternative check:Construct a model where one subset of rivers overlaps hills and a different subset overlaps deserts, with these subsets disjoint. All premises hold while II and III fail, confirming only I is compelled.Why Other Options Are Wrong:“All follow” and combinations including II or III assume overlaps not stated.“None follows” ignores the clear chain R ∩ D and D ⊆ Rd that yields I.Common Pitfalls:Illicitly merging different “some” groups; assuming that if two groups each overlap a third, they overlap each other.Final Answer:Only I follows

Question 7

Syllogism – Chain reasoning with “all” and “some” Statements: • Some saints are balls. • All balls are bats. • Some tigers are balls. Conclusions to test: I. Some bats are tigers. II. Some saints are bats. III. All bats are balls.

Accepted Answer

Introduction / Context:This item combines universal inclusion (“all balls are bats”) with existential overlaps (“some saints/tigers are balls”). The goal is to propagate membership through the universal set while testing which categorical conversions are valid.Given Data / Assumptions:Some saints are balls (S ∩ B ≠ ∅).All balls are bats (B ⊆ Bt).Some tigers are balls (T ∩ B ≠ ∅).Concept / Approach:If a particular element lies in “balls,” and “all balls are bats,” then that element automatically lies in “bats.” Therefore, any “some … are balls” statement converts into “some … are bats.” Conversely, “all bats are balls” is not supported by “all balls are bats” (that would be the illicit converse of a universal statement).Step-by-Step Solution:I. Some bats are tigers: Since some tigers are balls and every ball is a bat, those same tigers are bats. Hence I follows.II. Some saints are bats: Since some saints are balls and all balls are bats, those same saints are bats. Hence II follows.III. All bats are balls: The given universal is one-way (balls ⊆ bats). It does not establish the reverse (bats ⊆ balls). III does not follow.Verification / Alternative check:Imagine bats include many things besides balls; the premises remain true while III fails. I and II must hold due to direct subset propagation from B to Bt.Why Other Options Are Wrong:Options including III accept an illicit converse.“Only II follows” ignores the equally valid propagation for tigers.“None of these” is incorrect because I and II are compelled.Common Pitfalls:Conflating “all A are B” with “all B are A”; forgetting that “some A are B” can be pushed through a universal inclusion.Final Answer:Only I and II follow

Question 8

Syllogism – Books, schools, and colleges (careful with shared supersets) Statements: • Some pens are books. • All schools are books. • Some colleges are schools. Conclusions to test: I. Some colleges are pens. II. Some pens are schools. III. Some co…

Accepted Answer

Introduction / Context:This question illustrates a classic pitfall: when two classes are both contained in a larger class, we cannot infer that those two classes overlap each other. “All schools are books” and “some pens are books” do not guarantee any common element between schools and pens.Given Data / Assumptions:Some pens are books (P ∩ B ≠ ∅).All schools are books (S ⊆ B).Some colleges are schools (C ∩ S ≠ ∅).Concept / Approach:Use subset propagation carefully. From “some colleges are schools” and “all schools are books,” we can confidently move those specific colleges into the books set. However, moving from pens to schools (or colleges to pens) cannot be done without an explicit connecting statement beyond the shared superset “books.”Step-by-Step Solution:III: “Some colleges are books.” Since some colleges are schools and every school is a book, those colleges are books. III follows.I: “Some colleges are pens.” No link connects colleges to pens except both relate to books. Shared supersets do not imply overlap. I does not follow.II: “Some pens are schools.” Again, pens and schools both sit within books, but that does not force any intersection. II does not follow.Verification / Alternative check:Create a model where the “pens inside books” subset is entirely separate from the “schools (and thus colleges) inside books” subset. All premises hold; only III is true. Because no option lists “Only III,” the correct choice is “None of these.”Why Other Options Are Wrong:Any option pairing I or II asserts overlaps not supported by the premises.“All follow” overreaches; only III is compelled.Common Pitfalls:Assuming that two classes contained in the same larger class must overlap; ignoring that “some” statements identify only part of a set.Final Answer:None of these

Question 9

Syllogism – Buses, trains, rooms, and boats (universal negatives) Statements: • All trains are buses. • No room is a bus. • All boats are rooms. Conclusions to test: I. No boat is a train. II. No bus is a boat. III. No train is a room.

Accepted Answer

Introduction / Context:Here we combine two universal statements (“all” and “no”) to derive universal negatives. When a class is fully inside another (all trains ⊆ buses) and that other class is disjoint from a third (no rooms are buses), then the inside class is also disjoint from the third (no trains are rooms).Given Data / Assumptions:T ⊆ B (all trains are buses).R ∩ B = ∅ (no room is a bus).All boats are rooms (Bo ⊆ R).Concept / Approach:Use contrapositive-style reasoning with set relations. If class A is disjoint from B, then any subset of B is also disjoint from A. Also, if all X belong to a class disjoint from Y, then X and Y are disjoint.Step-by-Step Solution:III: From T ⊆ B and R ∩ B = ∅, it follows that T ∩ R = ∅, so “No train is a room.” III follows.I: Since Bo ⊆ R and R is disjoint from B, Bo is also disjoint from B. Because T ⊆ B, Bo ∩ T = ∅, hence “No boat is a train.” I follows.II: From Bo ⊆ R and R ∩ B = ∅, we have Bo ∩ B = ∅, so “No bus is a boat.” II follows.Verification / Alternative check:Draw three non-overlapping regions for buses and rooms, with trains placed entirely inside buses and boats entirely inside rooms. Visual inspection confirms all three conclusions.Why Other Options Are Wrong:Any option omitting one of the three conclusions contradicts the clean set-separation logic.“None follows” is clearly false because all three are mandated by the premises.Common Pitfalls:Forgetting that a subset of a set disjoint from another is also disjoint from that other; misreading “no A is B” as “some A are not B.”Final Answer:All follow

Question 10

Syllogism – Multiple “some” statements about mountains (avoid unjustified negatives) Statements: • Some mountains are hillocks. • Some mountains are rivers. • Some mountains are valleys. Conclusions to test: I. All mountains are either hillocks or r…

Accepted Answer

Introduction / Context:When all premises are “some”-type statements, they assert the existence of overlaps with a single set (“mountains”) but do not provide relationships among the non-mountain categories themselves. Any universal claim or cross-category relation beyond mountains must not be assumed.Given Data / Assumptions:M ∩ H ≠ ∅ (some mountains are hillocks).M ∩ R ≠ ∅ (some mountains are rivers).M ∩ V ≠ ∅ (some mountains are valleys).Concept / Approach:“Some” indicates at least one element exists in the intersection described. However, it does not cover all of M, nor does it establish any R–V, H–R, or H–V relations outside of M. Therefore, universal claims like “all mountains are …” or “no valley is a river” cannot be justified without additional data.Step-by-Step Solution:I: “All mountains are either hillocks or rivers or valleys.” The premises give three non-empty intersections but do not say anything about the remaining mountains. Some mountains could be none of the three. I does not follow.II: “No valley is river.” There is zero information about how V relates to R. II does not follow.III: “Some rivers are valleys.” Again, no given overlap between R and V is specified. III does not follow.Verification / Alternative check:Construct a model with three distinct mountain subgroups: M∩H, M∩R, and M∩V, all disjoint. All premises remain true while II and III fail and a portion of M lies outside H∪R∪V, refuting I.Why Other Options Are Wrong:Any option claiming one or more conclusions follow assumes extra relations not present.“Only either II or III” suggests an exclusive inference, which is unsupported.Common Pitfalls:Assuming that multiple “some” overlaps imply an overlap among the non-mountain categories; turning existential statements into universal coverage.Final Answer:None follows

Question 11

Syllogism – Three chained “some” statements (watch for non-guaranteed transitivity) Statements: • Some blades are hammers. • Some hammers are knives. • Some knives are axes. Conclusions to test: I. Some axes are hammers. II. Some knives are blades. …

Accepted Answer

Introduction / Context:A frequent trap in syllogisms is assuming that three separate “some” overlaps chain together to force an overlap between the first and the last categories. Without a single element proven to belong to all necessary intersections, no such transitive conclusion is guaranteed.Given Data / Assumptions:Bd ∩ Hm ≠ ∅ (some blades are hammers).Hm ∩ Kn ≠ ∅ (some hammers are knives).Kn ∩ Ax ≠ ∅ (some knives are axes).Concept / Approach:Each “some” might refer to different subsets of the middle category. For example, the hammers that are blades could be different from the hammers that are knives. Similarly, the knives that are hammers could be different from the knives that are axes. Without explicit linkage, you cannot assert overlaps like Ax ∩ Hm or Bd ∩ Kn.Step-by-Step Solution:I: “Some axes are hammers.” Not compelled; the knives that are axes need not be the same knives that are hammers.II: “Some knives are blades.” Not compelled; the hammers that are knives need not be the same hammers that are blades.III: “Some axes are blades.” This would require tracking an element across all three statements, which is not guaranteed.Verification / Alternative check:Construct a counterexample: pick distinct hammer subsets for the first two intersections and distinct knife subsets for the latter two; ensure no single element spans all. All premises remain true while I–III fail, proving none necessarily follow.Why Other Options Are Wrong:Options claiming any single conclusion follow assume a shared element across intersections that is not provided.“None of these” is inapplicable because the correct choice is explicitly “None follows.”Common Pitfalls:Forgetting that “some A are B” and “some B are C” does not imply “some A are C.”Final Answer:None follows

Question 12

Syllogism – Boxes, hammers, beads, rings (pushing existence through an “all”) Statements: • Some boxes are hammers. • Some hammers are beads. • All beads are rings. Conclusions to test: I. Some rings are hammers. II. Some hammers are boxes. III. Som…

Accepted Answer

Introduction / Context:Use the universal statement “all beads are rings” to move any known bead elements into the ring set. Combine that with “some hammers are beads” to produce a guaranteed overlap between rings and hammers. Separately, the “some boxes are hammers” statement converts to “some hammers are boxes.”Given Data / Assumptions:Bx ∩ Hm ≠ ∅ (some boxes are hammers).Hm ∩ Bd ≠ ∅ (some hammers are beads).Bd ⊆ Rg (all beads are rings).Concept / Approach:Push existence forward: elements in Bd are also in Rg. Therefore, the specific hammers that are beads are necessarily hammers that are rings. However, to claim “some rings are boxes,” we would need an element known to be both ring and box; the premises do not force that overlap.Step-by-Step Solution:I: Because some hammers are beads and all beads are rings, those elements are in Hm ∩ Rg. So “some rings are hammers” is true. I follows.II: From “some boxes are hammers,” we can rephrase equivalently as “some hammers are boxes.” II follows.III: “Some rings are boxes” would require a single element that is both a ring (via bead) and a box. The hammer that is a box may be different from the hammer that is a bead, so this need not hold. III does not follow.Verification / Alternative check:Build a model with disjoint subsets of hammers: one subset overlaps boxes; another subset overlaps beads (hence rings). All premises hold, but there is no ring–box overlap, confirming only I and II are guaranteed.Why Other Options Are Wrong:Any option including III assumes a shared element not established.“None follows” or “Only I follows” understates the valid inference in II.“All follow” overreaches.Common Pitfalls:Assuming two separate “some” statements must refer to the same individuals; forgetting valid reversal of a “some” statement (A ∩ B implies B ∩ A).Final Answer:Only I and II follow

Question 13

Syllogism – Blankets, beds, and pillows (using a universal to confirm existence) Statements: • Some blankets are beds. • Some pillows are blankets. • All beds are pillows. Conclusions to test: I. Some blankets are pillows. II. Some pillows are beds.…

Accepted Answer

Introduction / Context:Here, a universal statement (“all beds are pillows”) combined with an existential (“some blankets are beds”) guarantees existence inside the larger class (pillows). This allows multiple “some”-type conclusions to be derived safely.Given Data / Assumptions:Bl ∩ Bd ≠ ∅ (some blankets are beds).Pl ∩ Bl ≠ ∅ (some pillows are blankets).Bd ⊆ Pl (all beds are pillows).Concept / Approach:Any bed is a pillow. Therefore, the specific blankets that are beds are also pillows. In addition, because there exists at least one bed (from “some blankets are beds”), there exists at least one pillow that is a bed. Finally, the original “some blankets are beds” itself ensures some beds are blankets.Step-by-Step Solution:I: From Bl ∩ Bd ≠ ∅ and Bd ⊆ Pl, the same elements lie in Bl ∩ Pl. Thus “some blankets are pillows.” I follows.II: Since Bd ⊆ Pl and at least one bed exists (by Bl ∩ Bd), there is at least one pillow that is a bed. “Some pillows are beds” follows.III: “Some beds are blankets” is literally the first premise (Bl ∩ Bd ≠ ∅) rewritten, so III follows.Verification / Alternative check:Draw three sets with beds entirely inside pillows. Place at least one element in the overlap of blankets and beds. Instantly, all three conclusions can be read off the diagram.Why Other Options Are Wrong:Any option suggesting only one or two conclusions undervalues the chain from “some blankets are beds” and “all beds are pillows.”Exclusive “either/or” choices are inappropriate because all three hold simultaneously.Common Pitfalls:Overlooking that a universal joined with an existence statement yields existence in the superset; forgetting that restating a premise (“some beds are blankets”) is a valid derived conclusion.Final Answer:All I, II and III follow

Question 14

Syllogism – Dolls, bottles, cars, and windows (do not assume existence from universals) Statements: • All dolls are windows. • All bottles are windows. • All cars are bottles. Conclusions to test: I. All cars are windows. II. Some cars are dolls. II…

Accepted Answer

Introduction / Context:A key rule in syllogisms is that universal statements like “all A are B” do not by themselves guarantee the existence of any A. Conclusions that use “some” require at least one actual instance. We must therefore separate the safe subset propagation from any unjustified existential claims.Given Data / Assumptions:D ⊆ W (all dolls are windows).B ⊆ W (all bottles are windows).C ⊆ B (all cars are bottles).Concept / Approach:From C ⊆ B and B ⊆ W, we can derive C ⊆ W, i.e., “all cars are windows.” However, “some cars are dolls” and “some windows are cars” each require proof that at least one car exists. The premises never assert existence of cars; therefore such “some”-type conclusions are not logically forced.Step-by-Step Solution:I: All cars are windows: C ⊆ B and B ⊆ W imply C ⊆ W. The relationship holds universally but does not assert that any car actually exists.II: Some cars are dolls: This would need C ∩ D ≠ ∅. No statement links cars with dolls beyond both being inside windows, and existence of cars is not guaranteed. II does not follow.III: Some windows are cars: This needs C to be non-empty. Universals alone do not provide existence; thus III is not compelled.Verification / Alternative check:Consider a model with zero cars. All premises remain true, I remains a valid universal relation, but II and III fail due to lack of existence, proving they do not follow necessarily.Why Other Options Are Wrong:Any option that includes II or III assumes existence or extra overlap not provided.“Only I and III” is invalid because III lacks guaranteed existence.“All follow” is too strong.Common Pitfalls:Assuming existential import from universal premises; inferring overlap merely because two classes share a larger containing class.Final Answer:None of these

Question 15

Syllogism — Determine which conclusion(s) logically follow from the statements Statements: All tigers are lions. No cow is a lion. Some camels are cows. Conclusions: (I) Some lions are camels. (II) No camel is a tiger. (III) Some tigers are cows.

Accepted Answer

Introduction / Context:This is a standard syllogism (logical deduction) question. We are given three categorical statements about sets (tigers, lions, cows, camels) and asked which of the listed conclusions must be true. The task is to reason strictly from the given statements without importing external facts.Given Data / Assumptions:All tigers are lions (T ⊆ L).No cow is a lion (Cows ∩ L = ∅).Some camels are cows (∃ Camels ∩ Cows).We consider only necessary conclusions (must be true in every valid Venn arrangement consistent with the statements).Concept / Approach:If A ⊆ B and B is disjoint from C, then A is also disjoint from C.“Some X are Y” asserts existence of overlap; “No X is Y” asserts total disjointness.Test each conclusion by constructing or imagining Venn diagrams consistent with the premises.Step-by-Step Solution:From “Some camels are cows” and “No cow is a lion,” those particular camels are not lions.Since all tigers are lions, anything that is a tiger must be a lion. Therefore, anything not a lion cannot be a tiger.Thus, the camels that are cows (and hence non-lions) cannot be tigers. So Conclusion II (“No camel is a tiger”) follows.Conclusion I (“Some lions are camels”) is impossible because the camels we know about are cows, and cows have empty intersection with lions.Conclusion III (“Some tigers are cows”) is impossible because tigers are lions while cows are not lions, so their intersection is empty.Verification / Alternative check:Sketching a Venn diagram with L containing T entirely, and C disjoint from L, then placing some Camels overlapping C (outside L) confirms II and refutes I and III.Why Other Options Are Wrong:“Only I”, “Only III”, “Either I or II”, and “None” contradict the necessity just shown.Common Pitfalls:Assuming “some camels are cows” implies anything about camels and lions directly; always route through the disjointness and subset relations first.Final Answer:Only II follows

Question 16

Syllogism — Determine which conclusion(s) logically follow Statements: All flowers are toys. Some toys are trees. Some angels are trees. Conclusions: (I) Some angels are toys. (II) Some trees are flowers. (III) Some flowers are angels.

Accepted Answer

Introduction / Context:We must test which conclusions necessarily follow from three categorical statements about Flowers, Toys, Trees, and Angels. In syllogisms, a conclusion is valid only if it is true in every possible diagram that satisfies the statements.Given Data / Assumptions:All Flowers are Toys (F ⊆ T).Some Toys are Trees (∃ T ∩ Tr).Some Angels are Trees (∃ A ∩ Tr).We cannot assume extra overlaps unless forced by the statements.Concept / Approach:“All” gives subset relationships; “Some” asserts existence of overlap but not exclusivity.To show a conclusion does not necessarily follow, it is enough to construct a consistent model where the conclusion is false while all premises remain true.Step-by-Step Solution:Premises allow some Toys to be Trees, but those particular Toys need not be Flowers; Flowers are a subset of Toys, but may sit entirely away from the Trees segment.Similarly, “Some Angels are Trees” tells us Angels overlap Trees, but says nothing about overlapping Toys or Flowers.Thus, (I) “Some angels are toys” is not forced; Angels∩Trees could lie outside Toys.(II) “Some trees are flowers” also is not forced; the Trees∩Toys region can be disjoint from Flowers.(III) “Some flowers are angels” is not forced; Flowers (inside Toys) may have no overlap with Angels.Verification / Alternative check:Draw Trees intersecting Toys; place Flowers entirely in the Toys-but-not-Trees area; place Angels intersecting Trees but outside Toys. All premises hold; none of I, II, III hold—so none follows.Why Other Options Are Wrong:Any option claiming one or more conclusions follow adds overlaps not guaranteed by the premises.Common Pitfalls:Assuming transitivity through “Some.” Only “All” relationships chain; “Some” does not force further overlap.Final Answer:None follows

Question 17

Syllogism — Identify the valid conclusion(s) Statements: Some rats are cats. Some cats are dogs. No dog is a cow. Conclusions: (I) No cow is a cat. (II) No dog is a rat. (III) Some cats are rats.

Accepted Answer

Introduction / Context:This syllogism mixes “some” overlaps with a universal negative. We must decide which conclusions necessarily hold in every arrangement that satisfies the premises, without adding extra assumptions about further overlaps.Given Data / Assumptions:Some Rats are Cats (∃ R ∩ C).Some Cats are Dogs (∃ C ∩ D).No Dog is a Cow (D ∩ Cow = ∅).Concept / Approach:“Some X are Y” is symmetric, so “Some Cats are Rats” is equivalent to “Some Rats are Cats.”To invalidate a conclusion, it suffices to create a compliant diagram where that conclusion is false.Step-by-Step Solution:Conclusion III: “Some cats are rats.” This is exactly the first statement restated; it necessarily follows.Conclusion I: “No cow is a cat.” We only know Dogs are disjoint from Cows. Cats that are not Dogs could still be Cows; no statement forbids it. Hence I does not follow.Conclusion II: “No dog is a rat.” We know some Cats are Dogs and some Rats are Cats, but those two “some” groups need not be the same Cats. A model can keep R∩C and C∩D disjoint, so II does not follow.Verification / Alternative check:Draw Cats as a large set; place one overlap with Rats and a separate overlap with Dogs; ensure Dogs are disjoint from Cows. Premises hold; I and II can be false simultaneously while III holds.Why Other Options Are Wrong:Options including I or II assume extra exclusions not specified. Only III is compelled by the premises.Common Pitfalls:Treating two independent “some” overlaps as if they must coincide.Final Answer:Only III follows

Question 18

Syllogism — Determine which conclusion(s) must be true Statements: All tigers are jungles. No jungle is a bird. Some birds are rains. Conclusions: (I) No rain is a jungle. (II) Some rains are jungles. (III) No bird is a tiger.

Accepted Answer

Introduction / Context:Despite the odd wording, treat each term as a set label. We must determine which conclusions necessarily follow from the premises using set relationships and disjointness.Given Data / Assumptions:All Tigers are Jungles (T ⊆ J).No Jungle is a Bird (J ∩ B = ∅).Some Birds are Rains (∃ B ∩ R).Concept / Approach:If A ⊆ B and B is disjoint from C, then A is disjoint from C.“Some” statements do not force anything about the rest of the set outside the specified overlap.Step-by-Step Solution:From T ⊆ J and J ∩ B = ∅, it follows immediately that T ∩ B = ∅. So “No bird is a tiger” (III) must be true.Regarding (I) and (II): We only know some Birds are Rains. Those particular Birds are not Jungles, but other parts of Rains might still intersect (or not intersect) Jungles. The premises do not settle whether any Rain is a Jungle.Therefore neither “No rain is a jungle” (I) nor “Some rains are jungles” (II) is forced.Verification / Alternative check:Create two models: (a) Rains entirely within Birds; then I holds and II fails. (b) Split Rains into two parts, one overlapping Jungles outside Birds; then II holds and I fails. Since premises allow both models, neither I nor II necessarily follows, but III always does.Why Other Options Are Wrong:Options including I or II claim more than warranted. “All follow” is too strong.Common Pitfalls:Assuming the “some Birds are Rains” exhaustively describes Rains; it does not.Final Answer:Only III follows

Question 19

Syllogism — Identify the necessary conclusion(s) Statements: All snakes are trees. Some trees are roads. All roads are mountains. Conclusions: (I) Some mountains are snakes. (II) Some roads are snakes. (III) Some mountains are trees.

Accepted Answer

Introduction / Context:We are to deduce which conclusions must be true from the chain of subsets and an existential statement. Be careful not to overextend “some” beyond what is stated.Given Data / Assumptions:All Snakes are Trees (S ⊆ Tr).Some Trees are Roads (∃ Tr ∩ Rd).All Roads are Mountains (Rd ⊆ Mtn).Concept / Approach:Transitivity of “All”: if A ⊆ B and B ⊆ C, then A ⊆ C.“Some Trees are Roads” together with “Roads ⊆ Mountains” yields “Some Mountains are Trees.”But unless Snakes are directly linked to Roads, we cannot guarantee any Snake lies in Mountains via Roads.Step-by-Step Solution:From ∃(Tr ∩ Rd) and Rd ⊆ Mtn, we get ∃(Tr ∩ Mtn). Hence (III) “Some mountains are trees” necessarily follows.(I) “Some mountains are snakes” would require ∃(S ∩ Mtn). We only know S ⊆ Tr; we do not know that any Snakes are among the Trees which are also Roads. So (I) is not forced.(II) “Some roads are snakes” demands ∃(Rd ∩ S). We only have ∃(Rd ∩ Tr) and S ⊆ Tr, which does not guarantee any road-tree is a snake. Hence (II) does not follow.Verification / Alternative check:Construct a model where the Trees that are Roads are entirely distinct from the subset of Trees that are Snakes. Premises hold; (I) and (II) fail; (III) holds.Why Other Options Are Wrong:Options asserting (I) or (II) require extra overlap not provided by the premises.Common Pitfalls:Assuming that because S ⊆ Tr and some Tr are Rd, some S must be Rd. That is not logically necessary.Final Answer:Only III follows

Question 20

Syllogism — Determine which conclusion(s) follow Statements: All trees are flowers. No flower is a fruit. All branches are fruits. Conclusions: (I) Some branches are trees. (II) No fruit is a tree. (III) No tree is a branch.

Accepted Answer

Introduction / Context:This syllogism mixes universal affirmatives and negatives. We must identify conclusions that must be true in every valid interpretation of the premises and recognize mutually exclusive pairings where exactly one of two alternatives can hold.Given Data / Assumptions:All Trees are Flowers (T ⊆ F).No Flower is a Fruit (F ∩ Fr = ∅).All Branches are Fruits (B ⊆ Fr).Concept / Approach:From T ⊆ F and F disjoint from Fr, we get T ∩ Fr = ∅, i.e., no Fruit is a Tree (equivalently, no Tree is a Fruit).From B ⊆ Fr and T ∩ Fr = ∅, Trees and Branches are disjoint, so “No Tree is a Branch.”Hence of the pair (I) “Some branches are trees” and (III) “No tree is a branch,” exactly one can be true; given the premises, (I) is false and (III) is true.Step-by-Step Solution:Derive (II): T ⊆ F and F ∩ Fr = ∅ ⇒ T ∩ Fr = ∅. So “No fruit is a tree” must be true.Assess (I) vs (III): B ⊆ Fr and T ∩ Fr = ∅ ⇒ T ∩ B = ∅, which makes (I) false and (III) true. The option format groups them as “either I or III” along with (II); effectively, the valid combination is (III) and (II).Verification / Alternative check:Any Venn diagram respecting the premises places Trees entirely away from Fruits (and thus away from Branches). This makes (II) and (III) inevitable and contradicts (I).Why Other Options Are Wrong:Claiming only (II) ignores the forced disjointness with Branches (supporting III). Claiming (I) contradicts the established disjointness.Common Pitfalls:Missing that “No Flower is a Fruit” propagates through the subset “All Trees are Flowers,” eliminating any Tree–Fruit overlap.Final Answer:Only either I or III, and II follow

Logical Deduction Questions

Questions

Practice smarter