Question 1

Inference selection – Radios sold and quality standard: Statements: (i) All radios sold in that shop are of high standard. (ii) Some Murphy radios are sold in that shop. Inferences to test: I) All high-standard radios are manufactured by Murphy. II)…

Accepted Answer

Introduction / Context:We combine a universal statement about items sold at a shop with a particular about a brand (Murphy) to decide which inferences necessarily hold.Given Data / Assumptions:SoldInShop ⊆ HighStandard.Some Murphy ∩ SoldInShop (i.e., at least one Murphy radio is sold in the shop).Concept / Approach:Anything sold in the shop is high standard. Therefore those “some Murphy radios” that are sold there are high standard (II). Moreover, those same high-standard Murphy radios are indeed sold in the shop (IV). Inference I claims all high-standard radios are Murphy – an overreach. Inference III contradicts II.Step-by-Step Solution:From “Some Murphy are sold,” pick x ∈ Murphy ∩ SoldInShop.By the universal, x ∈ HighStandard.Hence “Some Murphy radios are of high standard” (II) and “Some high-standard Murphy radios are sold in that shop” (IV) both hold.Verification / Alternative check:No model can satisfy the premises and make II false; nor can it negate IV, since the specific x is both Murphy, high standard, and sold in that shop.Why Other Options Are Wrong:I is unsupported; III contradicts II.Common Pitfalls:Reversing subset directions or overlooking that “some” witnesses can be carried through universals.Final Answer:II and IV inferences only.

Question 2

Syllogism – Shared predicate without forced intersection: Statements: • All children are playful. • Some animals are playful. Conclusions: I) Some animals are children. II) Some children are animals.

Accepted Answer

Introduction / Context:Two sets (Children, Animals) share a common property (Playful) but no direct linkage is given between them. We must avoid inferring overlap merely from sharing a predicate.Given Data / Assumptions:Children ⊆ Playful.Some Animals are Playful (∃ Animal ∩ Playful).Concept / Approach:The premises do not assert any relation between Children and Animals, only that both touch Playful. Hence either overlap is possible or they may be disjoint; neither I nor II is forced.Step-by-Step Solution:Model A: Make Children and Animals disjoint inside Playful. Both premises hold; I and II fail.Model B: Make an overlap between Children and Animals; I and II may become true, but not by necessity.Verification / Alternative check:Necessity requires truth across all models; because countermodels exist, no conclusion follows.Why Other Options Are Wrong:They assume overlap not guaranteed by the premises.Common Pitfalls:“Same predicate fallacy”: treating a shared property as evidence of common identity.Final Answer:Neither I nor II follows.

Question 3

Syllogism – Some + No relation: Statements: (a) Some skirts are benches. (b) No bench is a table. Conclusions: I) Some skirts are tables. II) Some benches are skirts. III) All benches are skirts. IV) Some tables are skirts.

Accepted Answer

Introduction / Context:We combine one particular overlap with a universal exclusion to test which conclusions are necessary.Given Data / Assumptions:Some Skirts are Benches (∃ S ∩ B).No Bench is a Table (B ∩ T = ∅).Concept / Approach:Conclusion II restates the given particular in symmetric form and must hold. Conclusions I and IV claim a Skirt–Table overlap, which is not supported and in fact contradicted for the bench-portion of skirts by the “No bench is a table” clause. Conclusion III (“All benches are skirts”) is stronger than the “some” premise and is not compelled.Step-by-Step Solution:From ∃ S ∩ B, we can assert “Some benches are skirts” (II).Because B ∩ T = ∅, any element of B is not in T; therefore the overlap cannot imply any Skirt–Table overlap.Verification / Alternative check:Model: Let a few benches be skirts; keep tables disjoint from benches. The premises hold; II holds; I, III, IV do not necessarily follow.Why Other Options Are Wrong:They claim overlaps that are not guaranteed or universally quantify from a “some”.Common Pitfalls:Illicit conversion and over-generalizing from an existential.Final Answer:Only II follows.

Question 4

Syllogism – Universal inclusion with a universal exclusion: Statements: (a) All pens are pencils. (b) No pencil is a monkey. Conclusions: I) No pen is a monkey. II) Some pens are monkeys. III) All monkeys are pens. IV) Some monkeys are pens.

Accepted Answer

Introduction / Context:A chain of subset then disjointness produces a clean negative conclusion about Pens and Monkeys.Given Data / Assumptions:Pens ⊆ Pencils.Pencils ∩ Monkeys = ∅.Concept / Approach:If Pens sit inside Pencils, and Pencils are disjoint from Monkeys, then Pens and Monkeys are disjoint. That proves I. Claims II, III, IV contradict or go beyond this disjointness.Step-by-Step Solution:From Pens ⊆ Pencils and Pencils ∩ Monkeys = ∅, deduce Pens ∩ Monkeys = ∅.Therefore “No pen is a monkey” (I) is necessarily true.Verification / Alternative check:No countermodel can make a pen also a monkey while keeping “No pencil is a monkey” and “All pens are pencils”.Why Other Options Are Wrong:They contradict the disjointness or claim arbitrary either/or without support.Common Pitfalls:Missing the propagation of exclusion down the subset chain.Final Answer:Only conclusion I follows.

Question 5

Syllogism – Three chained universals: Statements: (a) All dogs are rats. (b) All rats are crows. (c) All crows are parrots. Conclusions: (I) All dogs are parrots. (II) Some parrots are dogs. (III) Some crows are dogs. (IV) All rats are dogs.

Accepted Answer

Introduction / Context:We have a pure chain of universals creating nested subsets: Dogs ⊆ Rats ⊆ Crows ⊆ Parrots. We must see which conclusions are necessary under standard aptitude conventions (which typically assume the subordinate sets are non-empty).Given Data / Assumptions:D ⊆ R ⊆ C ⊆ P.Standard test convention: the named sets are considered to have members unless negated.Concept / Approach:(I) follows directly from transitivity: D ⊆ P. (II) and (III) are particular statements that hold under the conventional assumption that D is non-empty (then some P are D; some C are D). (IV) reverses the first link and does not follow.Step-by-Step Solution:Chain the inclusions to get D ⊆ P (I).Because D is non-empty in typical exam readings, pick x ∈ D; then x ∈ P and x ∈ C, giving (II) and (III).(IV) “All rats are dogs” is the converse of (a) and is invalid.Verification / Alternative check:Diagram the nesting P ⊇ C ⊇ R ⊇ D. Choosing any dog yields a crow and a parrot simultaneously, validating (II) and (III) under the non-empty convention.Why Other Options Are Wrong:They either omit a valid consequence or include the false converse (IV).Common Pitfalls:Illicit converse; forgetting typical test convention about existence for named categories.Final Answer:Only (I), (II) and (III) follow.

Question 6

Syllogism – Two universals in a chain: Statements: (a) All animals are dogs. (b) All dogs are birds. Conclusions: I) All animals are birds. II) All birds are animals.

Accepted Answer

Introduction / Context:A simple transitive chain of universals yields one necessary inclusion; the converse does not hold.Given Data / Assumptions:Animals ⊆ Dogs ⊆ Birds.Concept / Approach:Transitivity produces Animals ⊆ Birds (I). The reverse “All birds are animals” adds information not provided and is not necessary.Step-by-Step Solution:From Animals ⊆ Dogs and Dogs ⊆ Birds, deduce Animals ⊆ Birds.Verification / Alternative check:Countermodel for II: Let Birds include animals and non-animals (e.g., mechanical birds); universals (a) and (b) can still hold while II fails.Why Other Options Are Wrong:They either include the invalid converse or deny the valid transitive inclusion.Common Pitfalls:Assuming symmetry of subset relations without basis.Final Answer:Only Conclusion I follows.

Question 7

Syllogism – One particular and one universal: Statements: • Some phones are watches. • All watches are guns. Conclusions: I) All guns are watches. II) Some guns are phones.

Accepted Answer

Introduction / Context:An existential (“some”) combined with a universal inclusion often yields a guaranteed “some” about the superset.Given Data / Assumptions:∃ Phone ∩ Watch.Watch ⊆ Gun.Concept / Approach:The particular witness that is both Phone and Watch must also be a Gun. Hence “Some guns are phones” (II) is necessary. Claim (I) reverses the universal and is not supported.Step-by-Step Solution:Pick x ∈ Phone ∩ Watch (exists).Since Watch ⊆ Gun, x ∈ Gun.Therefore x ∈ Phone ∩ Gun, proving II.Verification / Alternative check:No model can satisfy the premises and negate II, because the existential passes through the inclusion.Why Other Options Are Wrong:(I) claims Gun ⊆ Watch, which is not given.Common Pitfalls:Illicit conversion of universals.Final Answer:Only II follows.

Question 8

Syllogism – Universal about all students plus a particular about some students: Statements: (a) All students like excursions. (b) Some students like laboratory experiments. Conclusions: (I) Students who like laboratory experiments also like excursio…

Accepted Answer

Introduction / Context:We must test what is guaranteed when every student likes excursions and at least some like lab experiments.Given Data / Assumptions:Students ⊆ LikeExcursions.Some Students ⊆ LikeLab.Concept / Approach:Anyone who is a student (including those who like lab experiments) necessarily likes excursions by the universal. But (II) adds “some students do not like lab experiments,” which is not guaranteed by “some like lab.”Step-by-Step Solution:For (I): If x is a student and likes lab, x is a student; hence x likes excursions. So (I) follows.For (II): “Some like lab” does not imply “some do not like lab.” It remains possible that all students like lab; therefore (II) is not necessary.Verification / Alternative check:Model with every student liking both excursions and lab: premises true; (I) true; (II) false. Since (II) can fail, it is not necessary.Why Other Options Are Wrong:They include the non-necessary (II) or omit the necessary (I).Common Pitfalls:Reading “some” as “some but not all.” In logic, “some” allows “all”.Final Answer:Only (I) follows.

Question 9

Syllogism – Particular plus universal, reasoning about non-barking dogs: Statements: • Some dogs bark. • All dogs bite. Conclusions: I) Dogs that do not bark also bite. II) Dogs that do not bark do not necessarily bite.

Accepted Answer

Introduction / Context:We are asked to reason about the subset of dogs that do not bark, given that all dogs bite.Given Data / Assumptions:Some Dogs ⊆ Bark.Dogs ⊆ Bite (i.e., every dog bites).Concept / Approach:Since “bite” applies to all dogs, it also applies to every sub-group of dogs, including the non-barking ones. Therefore (I) is necessarily true and (II) is false.Step-by-Step Solution:Let NB = Dogs \ Bark (non-barking dogs). Because Dogs ⊆ Bite, NB ⊆ Bite as well.Hence every non-barking dog bites, proving (I).(II) contradicts this and is not necessary.Verification / Alternative check:No model can satisfy “All dogs bite” and produce a non-biting dog, barking or not.Why Other Options Are Wrong:They deny or weaken the universal inclusion.Common Pitfalls:Taking “some dogs bark” to restrict “all dogs bite” to only barking dogs.Final Answer:Only I follows.

Question 10

Syllogism – Two subsets of the same superset (Angels): Statements: (a) Some humane creatures are angels. (b) All doctors are angels. Conclusions: I) Some humane creatures are doctors. II) Some doctors are humane creatures.

Accepted Answer

Introduction / Context:We have two sets both lying inside Angels. Without a statement linking them, overlap is not guaranteed.Given Data / Assumptions:∃ Humane ∩ Angels.Doctors ⊆ Angels.Concept / Approach:(I) requires Humane ∩ Doctors ≠ ∅; (II) requires the same intersection but stated from the other side. Neither is forced purely from both being inside Angels.Step-by-Step Solution:Model A: Place Humane∩Angels disjoint from Doctors within Angels. Then I and II both fail while premises hold.Because countermodels exist, neither conclusion is necessary.Verification / Alternative check:Venn depiction with Angels as the universe, one region for Doctors, another for Humane∩Angels, non-overlapping.Why Other Options Are Wrong:They presuppose overlap not warranted by the premises.Common Pitfalls:Inferring intersection from shared superset membership.Final Answer:Neither I nor II follows.

Question 11

Syllogism – Universal negative: Statement: • No children are voters. Conclusions: I) All adults are voters. II) No voters are children.

Accepted Answer

Introduction / Context:From a universal negative about two sets (Children and Voters), we derive what is necessarily true about their relations.Given Data / Assumptions:Children ∩ Voters = ∅.No information about Adults is given.Concept / Approach:“No children are voters” is symmetric: it also means “No voters are children.” But it says nothing about adults; asserting that all adults are voters is unwarranted.Step-by-Step Solution:From C ∩ V = ∅, deduce V ∩ C = ∅ (same statement). Hence II follows.The statement provides zero linkage between Adults and Voters, so I is not forced.Verification / Alternative check:Construct models where many adults are voters, or few are voters; the premise remains true either way, proving I is not necessary while II is always true.Why Other Options Are Wrong:They either deny the symmetric negative or introduce unjustified universals about Adults.Common Pitfalls:Assuming “no children are voters” implies “all voters are adults”; the premise does not assert that.Final Answer:Only conclusion II follows.

Question 12

Syllogism — Peacocks, tigers, and lions (subset chaining):  Statements: (a) All peacocks are lions. (b) Some tigers are peacocks.  Conclusions to test: I. Some lions are not tigers. II. All tigers are lions. III. Some tigers are lions. IV. All peaco…

Accepted Answer

Introduction / Context:This problem checks your command over categorical syllogisms with subset chaining and the difference between “some” and “all.” We must mark only the conclusions that are logically necessary under every model that satisfies the statements.Given Data / Assumptions:(a) All peacocks are lions (Peacock ⊆ Lion).(b) Some tigers are peacocks (∃ Tiger ∩ Peacock).Standard test convention: existence is guaranteed only where explicitly stated (“some”).Concept / Approach:Use set inclusion and intersection. If some tigers are peacocks and every peacock is a lion, those specific tigers are lions. No other universal or negative relation is licensed by the premises.Step-by-Step Solution:1) From (b), pick at least one entity that is both Tiger and Peacock.2) From (a), Peacock ⊆ Lion, so that entity is also a Lion.3) Therefore, ∃ (Tiger ∩ Lion): “Some tigers are lions” (III) must be true.4) I “Some lions are not tigers” is not forced; all lions could, in principle, be tigers.5) II “All tigers are lions” generalizes far beyond the premises and is not warranted.6) IV “All peacocks are tigers” reverses the inclusion direction and is unsupported.Verification / Alternative check:Create a model: Peacocks form a small subset inside Lions; some Tigers overlap that Peacock subset. III is compelled; I, II, and IV can vary without violating the premises.Why Other Options Are Wrong:Each of I, II, IV adds information not entailed by (a) and (b).Common Pitfalls:Assuming “some” implies “all,” or reversing subset directions.Final Answer:Only conclusion III follows.

Question 13

Syllogism — Mobiles, watches, and calculators (careful with middle term):  Statements: (a) No mobile is a watch. (b) All watches are calculators.  Conclusions to test: I. No calculator is a mobile. II. Some calculators are mobiles.  Select what nece…

Accepted Answer

Introduction / Context:This question tests whether you can avoid illicit conversions when a middle term changes grammatical role. “Watch” is the bridge between “mobile” and “calculator,” but we must not infer more than the premises license.Given Data / Assumptions:No Mobile is a Watch (Mobile ∩ Watch = ∅).All Watch are Calculator (Watch ⊆ Calculator).No explicit existence of watches is stated.Concept / Approach:If there are watches, they fall inside calculators and are disjoint from mobiles. That says nothing about other calculators (non-watch calculators) possibly being or not being mobiles. Hence neither a universal negative (I) nor an existential positive (II) about the class Calculator vs Mobile is forced.Step-by-Step Evaluation:1) From (b), some calculators could be watches (if watches exist), and those are certainly not mobiles (by (a)).2) But calculators may also include devices that are not watches; the premises give no restriction on whether those could be mobiles.3) Therefore, “No calculator is a mobile” (I) is too strong, and “Some calculators are mobiles” (II) is not guaranteed.Verification / Alternative check:Model A: calculators = watches only → I would be true; Model B: calculators include some non-watch mobiles → II would be true. Since different models satisfy different conclusions, neither is necessary.Common Pitfalls:Confusing “All watches are calculators” with “All calculators are watches,” or assuming the existence of watches.Final Answer:Neither Conclusion I nor II follows.

Question 14

Syllogism — Benches, tables, and chairs (direction of inclusion):  Statements: (a) All benches are tables. (b) No table is a chair.  Conclusions to test: I. All benches are chairs. II. All tables are benches. III. No tables are benches. IV. No bench…

Accepted Answer

Introduction / Context:Here we combine a universal inclusion with a universal exclusion. The goal is to respect the direction of inclusion and not convert statements illicitly.Given Data / Assumptions:Bench ⊆ Table.Table ∩ Chair = ∅.Concept / Approach:If all benches are tables and no table is a chair, then benches—being tables—cannot be chairs. Other conclusions either reverse the inclusion or contradict the premises.Step-by-Step Assessment:1) I “All benches are chairs” contradicts Table ∩ Chair = ∅ because benches are tables.2) II “All tables are benches” reverses Bench ⊆ Table and is not implied.3) III “No tables are benches” is false; benches are a subset of tables, so that intersection is non-empty if benches exist.4) IV “No benches are chairs” follows immediately from benches ⊆ tables and tables disjoint from chairs.Common Pitfalls:Conflating “All A are B” with “All B are A,” and ignoring that a subset inherits all disqualifications that apply to the superset.Final Answer:Only conclusion IV follows.

Question 15

Syllogism — Books, calculators, and mobiles (reading “some” correctly):  Statements: (a) Some books are mobiles. (b) Some calculators are mobiles.  Conclusions to test: I. Some mobiles are calculators. II. Some mobiles are books.

Accepted Answer

Introduction / Context:This is a straightforward test of interpreting “some” statements. Each premise explicitly asserts a non-empty overlap with the class “mobiles.”Given Data / Assumptions:∃ (Book ∩ Mobile).∃ (Calculator ∩ Mobile).Concept / Approach:From the premises themselves we can restate the conclusions: “Some mobiles are books” is equivalent to “Some books are mobiles,” and “Some mobiles are calculators” is equivalent to “Some calculators are mobiles.” Both must be true.Step-by-Step:1) Use premise (a) to assert existence of at least one entity that is both Book and Mobile → conclusion II.2) Use premise (b) to assert existence of at least one entity that is both Calculator and Mobile → conclusion I.Verification:A Venn diagram with Mobile as the central set and two overlapping “pockets” for Books and Calculators satisfies both conclusions immediately.Common Pitfalls:Overthinking “some” as implying additional relations between Books and Calculators; none are needed for these conclusions.Final Answer:Both conclusions I and II follow.

Question 16

Syllogism — Stones, gold, and banks (don’t force extra overlap):  Statements: (a) Some stones are gold. (b) Some gold are banks.  Conclusions to test: I. Some banks are stones. II. Some gold are stones. III. No stone is bank. IV. Some banks are gold.

Accepted Answer

Introduction / Context:The premises provide two separate “some” relationships that share the class “gold.” The trap is to assume those “some” subsets are the same individuals, which is not required.Given Data / Assumptions:∃ (Stone ∩ Gold).∃ (Gold ∩ Bank).Concept / Approach:Statements with “some” guarantee existence of at least one element in the intersection but do not identify which elements. Therefore, the overlap between Stone and Bank is undetermined, while the two direct conclusions about the given overlaps must hold.Step-by-Step Assessment:1) II “Some gold are stones” is exactly premise (a) restated and hence follows.2) IV “Some banks are gold” is exactly premise (b) restated and hence follows.3) I requires the two “some” groups to coincide; that is not necessary.4) III contradicts premise (b) if banks exist; in any case, it is not compelled by the premises.Common Pitfalls:Forcing a transitive “some” across two intersections or assuming disjointness without evidence.Final Answer:Only conclusions II and IV follow.

Question 17

Syllogism — Men, bachelors, teachers, and spiritual (chain with partial existence):  Statements: (a) All men are bachelors. (b) Some bachelors are teachers. (c) Some teachers are spiritual.  Conclusions to test: I. All men are spiritual. II. Some te…

Accepted Answer

Introduction / Context:This item mixes a universal inclusion with two existential premises. Be careful not to overgeneralize from “some,” and avoid reversing subset relations.Given Data / Assumptions:Man ⊆ Bachelor.∃ (Bachelor ∩ Teacher).∃ (Teacher ∩ Spiritual).Concept / Approach:We know that at least one Teacher is Spiritual (from (c)); that alone makes II true. But nothing guarantees that any Man is among those Spiritual Teachers, since the “some bachelors are teachers” group might be disjoint from the Men subset within Bachelors.Step-by-Step Evaluation:1) From (c), immediately, “Some teachers are spiritual” (II) follows.2) I “All men are spiritual” is far too strong; neither (b) nor (c) universalizes across teachers, let alone men.3) III “Some men are spiritual” would require explicit linkage between Men and the Spiritual Teachers; not provided.4) IV “All teachers are spiritual” contradicts the “some” nature of (c).Common Pitfalls:Assuming that because all men are bachelors, they are automatically part of the specific bachelor-teacher group named in (b); “some” does not imply inclusion of every subset.Final Answer:Only conclusion II follows.

Question 18

Syllogism — Food, sweet, and sour (reading independence of “some”):  Statements: • Some food are sweet. • Some food are sour.  Conclusions to test: (I) All food are either sweet or sour. (II) Some sweets are sour.

Accepted Answer

Introduction / Context:This is a classic test of not over-interpreting two particular (“some”) statements. Each premise speaks about a possibly different subset of “food.”Given Data / Assumptions:∃ (Food ∩ Sweet).∃ (Food ∩ Sour).Concept / Approach:Because “some” does not mean “all,” the class Food may contain items that are neither sweet nor sour; likewise, the sweet-food items and sour-food items could be entirely different individuals, so there is no guarantee of overlap between Sweet and Sour.Step-by-Step Evaluation:1) (I) “All food are either sweet or sour” is a universal claim and cannot be inferred from two existential premises.2) (II) “Some sweets are sour” would require a non-empty intersection Sweet ∩ Sour; the premises permit this to be empty because the “some” groups might be disjoint.Verification / Alternative check:Construct a model with 10 foods: 3 are sweet, 3 are sour, 4 are neither; the premises hold, yet neither (I) nor (II) is forced.Common Pitfalls:Assuming that two “some” statements imply overlap or exhaustiveness.Final Answer:Neither Conclusion (I) nor (II) follows.

Question 19

Syllogism — Ladies, beautiful, honest, sensitives (multi-step reasoning):  Statements: • Some ladies are beautiful. • Some beautiful are honest. • All honest are sensitives.  Conclusions: 1) Some sensitives are beautiful. 2) Some honest are ladies. …

Accepted Answer

Introduction / Context:This question chains one existential and one universal premise. The central caution is that “some” subsets named in different premises need not refer to the same individuals unless explicitly connected.Given Data / Assumptions:∃ (Lady ∩ Beautiful).∃ (Beautiful ∩ Honest).Honest ⊆ Sensitive.Concept / Approach:From “Some Beautiful are Honest” and “All Honest are Sensitives,” we can conclude that at least one entity is both Beautiful and Sensitive. That certifies Conclusion 1. However, nothing requires any Lady to be among those particular Honest or Sensitive individuals.Step-by-Step Evaluation:1) Take an element from Beautiful ∩ Honest (guaranteed by the second premise).2) Map it via Honest ⊆ Sensitive to get Beautiful ∩ Sensitive — proving Conclusion 1.3) Conclusion 2 would require an overlap between Honest and Lady; this is not guaranteed by the premises.4) Conclusion 3 would require an overlap between Sensitive and Lady; again, not guaranteed.Common Pitfalls:Conflating separate “some” groups or assuming transitive overlap among existentials without evidence.Final Answer:Only Conclusion 1 follows.

Question 20

Syllogism — Pens, pencils, and erasers (universal negative via subset):  Statements: • All pens are pencils. • No pencil is an eraser.  Conclusions: I. No eraser is a pen. II. No pen is an eraser.

Accepted Answer

Introduction / Context:We combine a subset relation with a universal exclusion. The key is to see that exclusions apply to all subsets of the excluded class.Given Data / Assumptions:Pen ⊆ Pencil.Pencil ∩ Eraser = ∅.Concept / Approach:If nothing in Pencil is an Eraser, then no Pen (being a Pencil) can be an Eraser. The converse statement “No eraser is a pen” also follows because if an Eraser were a Pen, it would be a Pencil, contradicting the exclusion.Step-by-Step:1) From Pen ⊆ Pencil and Pencil ∩ Eraser = ∅, infer Pen ∩ Eraser = ∅ → II follows.2) Suppose, for contradiction, an Eraser is a Pen. Then it is a Pencil, violating Pencil ∩ Eraser = ∅ → I follows.Common Pitfalls:Believing only one direction holds; here both non-membership statements are equivalent because of the subset.Final Answer:Both conclusions I and II follow.

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