Question 1

Calendars — On which day of the week is Arti's birthday? Statements: I. Arti's birthday is after Wednesday but before Sunday. II. Arti's birthday is after Friday but before Tuesday.

Accepted Answer

Introduction / Context:We must uniquely identify a weekday using two range constraints.Given Data / Assumptions:Week sequence: Mon, Tue, Wed, Thu, Fri, Sat, Sun.I: Day ∈ {Thu, Fri, Sat}.II: Day ∈ {Sat, Sun, Mon}.Concept / Approach:Intersect the candidate sets from both statements.Step-by-Step Solution:From I, candidates are Thu/Fri/Sat.From II, candidates are Sat/Sun/Mon.Intersection = {Sat}. Hence, the birthday must be on Saturday.Verification / Alternative check:No other day satisfies both inequalities simultaneously; the intersection is a singleton.Why Other Options Are Wrong:I alone: three possibilities remain.II alone: three possibilities remain.Either alone sufficient: false.Both together not sufficient: false.Common Pitfalls:Misreading boundaries ('after' excludes the day mentioned; 'before' excludes the endpoint).Final Answer:Both together are sufficient (Saturday).

Question 2

Coding — What is the code for 'sure'? Statements: I. 'he is sure' → 'ja ha main'. II. 'is she sure' → 'ka ja main'.

Accepted Answer

Introduction / Context:Identify the unique codeword for 'sure' from two coded sentences.Given Data / Assumptions:S1: 'he is sure' ↔ {ja, ha, main}.S2: 'is she sure' ↔ {ka, ja, main}.Concept / Approach:Common English words across the sentences are {is, sure}. The common codes are {ja, main}. Without an additional sentence breaking the tie, we cannot tell which of {ja, main} maps to 'sure' and which to 'is'.Step-by-Step Solution:From S1 alone: three words, three codes — but mapping is not fixed.From S2 alone: same issue.Together: intersection yields two possible assignments: (sure→ja, is→main) or (sure→main, is→ja). Both satisfy the data.Verification / Alternative check:Try to use the unique words (he ↔ ha in S1; she ↔ ka in S2). These do not help distinguish between 'ja' and 'main' for 'sure'.Why Other Options Are Wrong:I alone / II alone: insufficient because multiple bijections exist.Either alone sufficient / Both together sufficient: contradicted by two valid mappings.Common Pitfalls:Assuming alphabetical or positional mapping; presuming 'main' must mean 'is' due to meaning (semantics are irrelevant in this puzzle).Final Answer:Both Statements I and II together are not sufficient.

Question 3

Ages — Among P, Q, R, S, T (all different ages), who is the youngest? Statements: I. Q is younger than only P. II. S is older than only R.

Accepted Answer

Introduction / Context:We must find the youngest person using partial order statements that may even conflict.Given Data / Assumptions:I: 'Q is younger than only P' ⇒ only P is younger than Q, so P is the youngest and Q is second youngest.II: 'S is older than only R' ⇒ only R is younger than S, so R is the youngest and S is second youngest.Concept / Approach:Check consistency; if I implies youngest=P and II implies youngest=R, then the two cannot simultaneously be true unless P=R (not allowed).Step-by-Step Solution:From I alone: youngest is P (unique).From II alone: youngest is R (unique).Taken together, there is a contradiction regarding who is youngest. Since the task is to answer using the given statements (which must both hold if used together), the joint use does not yield a consistent unique answer.Verification / Alternative check:If we try to reconcile by reinterpreting phrasing, we would be adding assumptions. The DS framework disallows that.Why Other Options Are Wrong:I alone sufficient / II alone sufficient: each is sufficient individually, but the question's options are about sufficiency classification; contradiction between them means we cannot pick 'either alone' as sufficient to the DS prompt of which option describes sufficiency generally.Both together sufficient: false due to inconsistency.Common Pitfalls:Overlooking that contradictory statements make the pair unusable; misreading 'younger than only P' vs 'older than only R'.Final Answer:Both Statements I and II together are not sufficient.

Question 4

Calendars — On which day did Sourav visit Delhi? Statements: I. Sourav visited after Monday but before Thursday, and not on an odd day of the week. II. Sourav visited before Friday but after Monday.

Accepted Answer

Introduction / Context:Identify a weekday from range and parity constraints.Given Data / Assumptions:Week sequence assumed: Mon(1), Tue(2), Wed(3), Thu(4), Fri(5), Sat(6), Sun(7).I: After Monday and before Thursday, not on an odd day ⇒ candidates from {Tue(2), Wed(3)} but excluding odd ⇒ Tuesday only.II: After Monday and before Friday ⇒ {Tue, Wed, Thu} (three possibilities).Concept / Approach:Use parity information in I to isolate a single day.Step-by-Step Solution:From I alone: Tuesday is uniquely determined.From II alone: three possible days remain; insufficient.Verification / Alternative check:If one numbers days differently, parity could change; however, the conventional numbering (Mon=1) is standard for such DS problems and is implied by the typical 'odd/even day' phrasing.Why Other Options Are Wrong:II alone: multiple candidates.Either alone / Both together not sufficient / Both together sufficient: inconsistent with the unique determination by I.Common Pitfalls:Including Thursday in I's range ('before Thursday' excludes Thursday); misinterpreting 'odd day' without a day index convention.Final Answer:Statement I alone is sufficient; Statement II alone is not sufficient. (Tuesday)

Question 5

Seating (Left–Right positions) — What is R's position from the left end in a row? Statements: I. M is tenth from the left end of the row. II. There are sixteen children between M and R.

Accepted Answer

Introduction / Context:We must find R's exact position from the left end using relative placement information. Data Sufficiency (DS) requires checking if a unique answer is implied without assuming missing facts like total class size unless forced by logic.Given Data / Assumptions:I: M is 10th from the left.II: There are 16 children between M and R.Positions are whole numbers; a person cannot be at position ≤ 0.Concept / Approach:Translate the 'k between' language into a fixed offset of k+1 places to the left or right. Validate side choices against boundary feasibility.Step-by-Step Solution:From I alone: R's position is unknown; we only know M = 10th.From II alone: Only the gap between M and R is known (16 between), but neither absolute side nor any anchor position is given.Combine I & II: If R is to the right of M, then R = 10 + 16 + 1 = 27th. If R is to the left, R = 10 − 16 − 1 = −7 (impossible). Hence only the right-side placement is valid ⇒ R is 27th from the left.Verification / Alternative check:We do not need the row length; the invalid negative index eliminates the left-side case, leaving a unique feasible position.Why Other Options Are Wrong:A/B/C: Each single statement lacks enough information.D: Incorrect, because together they yield a single valid position.Common Pitfalls:Forgetting to add +1 when converting 'k between' to the actual offset; overlooking boundary invalidation on the left side.Final Answer:E — Together sufficient (R is 27th from the left).

Question 6

Ranks — What is Suneeta's rank from the top in a class of students? Statements: I. In a class of 42 students, Suneeta is 29th from the bottom. II. Suneeta is ten ranks below Samir.

Accepted Answer

Introduction / Context:This is a rank-conversion problem: converting 'from bottom' to 'from top' when class size is known.Given Data / Assumptions:Total students = 42 (from I).Suneeta is 29th from bottom (from I).From II: Suneeta is ten ranks below Samir, but Samir’s rank is unknown.Concept / Approach:Use the identity: rank_from_top = N − rank_from_bottom + 1.Step-by-Step Solution:From I: rank_from_top = 42 − 29 + 1 = 14. Unique.From II alone: relative to Samir only; cannot compute an absolute rank.Verification / Alternative check:Plugging back: 14th from top implies 29th from bottom in a class of 42, which matches I.Why Other Options Are Wrong:B: II lacks absolute anchor.C: Neither statement alone other than I suffices.D/E: Unnecessary; I already gives a unique answer.Common Pitfalls:Forgetting the +1 while converting between top/bottom ranks.Final Answer:A — Statement I alone suffices (rank = 14th from the top).

Question 7

Calendars (Leap day) — On which year (day is Feb 29) was Yasir born? Statements: I. He was born between years 2005 and 2011. II. He will complete 4 years on February 29, 2012.

Accepted Answer

Introduction / Context:We must identify the birth year for a Feb 29 birthday using leap-year logic and a completion-age cue.Given Data / Assumptions:Yasir's birthday is Feb 29.I: Birth year ∈ {2006, 2007, 2008, 2009, 2010}. Among these, only 2008 is a leap year.II: He completes 4 years on Feb 29, 2012.Concept / Approach:Completion of 4 years on the same calendar date implies birth 4 years earlier on that date.Step-by-Step Solution:From II: Feb 29, 2012 minus 4 years = Feb 29, 2008 → unique.From I alone: The interval narrows to 2008 by leap-year filtering, but DS convention avoids injecting external knowledge unless inherent; still, without the ”completes 4 years on” anchor, I is not enough by itself to prove uniqueness to all test-takers.Verification / Alternative check:2012 and 2008 are both leap years; the 4-year gap is consistent.Why Other Options Are Wrong:A/C/E/D: II alone already identifies 2008; no combination is needed.Common Pitfalls:Confusing 'turns 4 on' with 'has completed 4 years sometime before'; the exact phrasing pins the date.Final Answer:B — Statement II alone suffices (Birth year = 2008).

Question 8

Sets (Games) — Out of 64 students, 38 play both chess and cricket. How many play only chess? Statements: I. Of 64 students, 22 play no game; 4 play only cricket. II. Of 64 students, 20 are girls and 10 girls play no game.

Accepted Answer

Introduction / Context:Classic two-set (chess, cricket) problem asking for the 'only chess' count.Given Data / Assumptions:Total students = 64.Both chess & cricket = 38.I: None = 22; Only cricket = 4.II: Girl counts (irrelevant to chess/cricket split unless further data is given).Concept / Approach:Use inclusion–exclusion on the playing subset.Step-by-Step Solution:From I: Number who play at least one game = 64 − 22 = 42.Let x = only chess. Then x + (only cricket 4) + (both 38) = 42 ⇒ x = 0.Thus, 'only chess' = 0 (unique).From II alone: Gender information does not connect to game partitions; cannot solve.Verification / Alternative check:Check non-negativity and totals: 0 + 4 + 38 + 22 = 64 ✔.Why Other Options Are Wrong:B/C/D/E: II alone is irrelevant; I alone already suffices.Common Pitfalls:Forgetting that 'k between' style offsets do not apply here; misapplying inclusion–exclusion.Final Answer:A — Statement I alone suffices (Only chess = 0).

Question 9

Totals (Growth rates) — What is the total number of students in the school this year? Statements: I. The ratio of girls to boys is 2:3. II. The number of students grew by 5% this year and by 4% last year, starting from 2000 the year before last.

Accepted Answer

Introduction / Context:We must compute this year’s total from chained percentage growth starting from a known baseline.Given Data / Assumptions:Year before last: 2000 students.Last year growth: +4%.This year growth: +5% (on last year).Gender ratio 2:3 is irrelevant for totals.Concept / Approach:Apply successive percentage growth multiplicatively.Step-by-Step Solution:Last year = 2000 * (1 + 4/100) = 2000 * 1.04 = 2080.This year = 2080 * (1 + 5/100) = 2080 * 1.05 = 2184.Verification / Alternative check:Order of application matches chronology; no rounding needed for an integer cohort when base is integral in this setup.Why Other Options Are Wrong:A: A ratio without any count cannot yield a total.C/D/E: Statement II alone already yields a unique total.Common Pitfalls:Adding percentages instead of compounding; mixing base years.Final Answer:B — Statement II alone suffices (Total = 2184 students).

Question 10

Heights — Who among six (Geetha, Shipa, Deepa, Meeta, Sunita, Sadhana) is the tallest? Base facts: Geetha > Shipa; Deepa > Meeta. Statements: I. Sadhana > Sunita. II. Sadhana > Shipa and Meeta and Deepa.

Accepted Answer

Introduction / Context:Determine the tallest using partial orderings among six people. DS checks whether a unique maximal element is forced.Given Data / Assumptions:Base: Geetha > Shipa; Deepa > Meeta.I: Sadhana > Sunita.II: Sadhana > Shipa, Meeta, Deepa.Concept / Approach:Consolidate known relations and see whether anyone is shown to dominate all others.Step-by-Step Solution:From II: Sadhana is taller than Shipa, Meeta, and Deepa. From base, Geetha > Shipa (but no relation between Geetha and Sadhana).I only adds Sadhana > Sunita. After combining I+II+base, we still lack a comparison between Sadhana and Geetha.Thus, either Sadhana is tallest (if Sadhana > Geetha) or Geetha is tallest (if Geetha > Sadhana). Both possibilities satisfy all statements.Verification / Alternative check:Construct models: (1) Sadhana > Geetha > Deepa > Meeta > Shipa > Sunita; (2) Geetha > Sadhana > Deepa > Meeta > Shipa > Sunita. Both are consistent.Why Other Options Are Wrong:A/B/C/E: No single statement or their combination determines Sadhana vs Geetha.Common Pitfalls:Assuming transitive links that are not stated (e.g., inferring Geetha > Deepa without basis).Final Answer:D — Together not sufficient (tallest could be Sadhana or Geetha).

Question 11

Calendars — On which date is Amit's birthday in September 2010? Statements: I. In September 2009, his birthday fell on the last Thursday of the month. II. In September 2010, his birthday will fall on the last Friday of the month.

Accepted Answer

Introduction / Context:We need the exact date (calendar date number) of Amit's September 2010 birthday using weekday placement information.Given Data / Assumptions:II provides that in Sep 2010 the birthday is on the last Friday.Month/year are fixed (September 2010 has 30 days).I references Sep 2009, which is not required if II already pins the weekday position within Sep 2010.Concept / Approach:For a fixed month/year, the phrase 'last Friday' corresponds to a unique date.Step-by-Step Solution:Determine the dates for Fridays in September 2010 and select the largest one within 1–30. (This yields a unique date number.)Statement I is historical and does not alter September 2010’s weekday alignment for identifying the last Friday.Verification / Alternative check:Calendar enumeration confirms a single 'last Friday' date exists for any specified month/year.Why Other Options Are Wrong:A/C/E/D: II alone suffices; I is superfluous.Common Pitfalls:Thinking the previous year's weekday is required; it is not when the target month/year is explicit.Final Answer:B — Statement II alone suffices (unique 'last Friday' date in Sep 2010).

Question 12

Weights — Among M, P, K, J, T, W, who is lighter than only the heaviest (i.e., second heaviest)? Statements: I. P is heavier than M and T. II. W is heavier than P but lighter than J, and J is not the heaviest.

Accepted Answer

Introduction / Context:We need the person who is lighter than only the heaviest (rank 2 by weight). Partial inequalities are given.Given Data / Assumptions:I: P > M and P > T.II: J is above W (J > W) and W > P; additionally, J is not the heaviest, meaning someone else is heavier than J.No relations given for K against others.Concept / Approach:Combine inequalities; check if any person is forced to be second heaviest.Step-by-Step Solution:From II: There exists some X with X > J > W > P. From I: P > M, T. K remains unconstrained.Second-heaviest candidate could be J (if only X > J and all others < J), but K might also exceed J because we lack K's relation; in that case, second-heaviest could be K or even X depending on structure.Therefore, no unique identification is possible.Verification / Alternative check:Construct consistent orders where the second heaviest differs (e.g., Case A: X > J > W > P > K > M > T → second = J; Case B: X > K > J > W ... → second = K).Why Other Options Are Wrong:A/B/C/E: Insufficient constraints; both statements together still leave K and the unknown heaviest free.Common Pitfalls:Assuming transitivity to absent elements; overlooking that 'J not heaviest' does not place J relative to K.Final Answer:D — Together not sufficient.

Question 13

Coding — How is 'never' written in the code language? Statements: I. 'never ever go there' → 'na ja ni ho'. II. 'go there and come back' → 'ma ho sa ni da'.

Accepted Answer

Introduction / Context:We must identify the unique code token for the English word 'never' via overlapping coded sentences.Given Data / Assumptions:S1: 'never ever go there' ↔ {na, ja, ni, ho}.S2: 'go there and come back' ↔ {ma, ho, sa, ni, da}.One-to-one mapping per sentence; tokens are word-level.Concept / Approach:Intersect common words and codes to eliminate shared pairs; leftover tokens in S1 would correspond to {never, ever} but remain ambiguous.Step-by-Step Solution:Common English words across S1 and S2: {go, there}. Common codes: {ni, ho} (order unknown).Removing these from S1 leaves {na, ja} mapping to {never, ever}, which cannot be distinguished without an additional sentence containing exactly one of {never, ever}.Verification / Alternative check:No arrangement of the S2 tokens resolves the tie between {na, ja}.Why Other Options Are Wrong:A/B/C/E: Neither statement alone nor both together can uniquely assign 'never'.Common Pitfalls:Assuming alphabetical/semantic hints; DS forbids using such external cues.Final Answer:D — Together not sufficient.

Question 14

Symbolic language — What does 'S' mean? Statements: I. '5 S # 3' means 'flowers are really good'. II. '7 # 3 5' means 'flowers are available'.

Accepted Answer

Introduction / Context:Determine the English meaning of the symbol 'S' from two coded statements.Given Data / Assumptions:S1 tokens {5, S, #, 3} ↔ {flowers, are, really, good} (order unknown).S2 tokens {7, #, 3, 5} ↔ {flowers, are, available} (four tokens for three words implies one token corresponds to a word common with S1; exact mapping still unknown).Concept / Approach:Use overlaps: common English words across S1 and S2 are {flowers, are}. Common tokens across sentences are {5, #, 3}. Thus, these three tokens collectively map to two words; ambiguity remains.Step-by-Step Solution:Since S appears only in S1, it must map to one of {really, good}. However, S1 alone cannot decide which; S2 does not include either of those words, so it cannot resolve the tie.Verification / Alternative check:Any attempt to force S = 'really' or S = 'good' conflicts with the bijection constraints without additional evidence.Why Other Options Are Wrong:A/B/C/E: The pair of statements does not uniquely pin 'S' to a single word.Common Pitfalls:Assuming token positions correspond to word order; DS problems typically treat them as sets.Final Answer:D — Together not sufficient.

Question 15

Blood relations — How is P related to J? Statements: I. M is the brother of P; T is the sister of P. II. P's mother is married to J's husband, who has one son and two daughters.

Accepted Answer

Introduction / Context:We must find P's relation to J from partial family information that allows multiple consistent family trees.Given Data / Assumptions:I: Sibling facts about P (one brother, one sister mentioned), but nothing about J.II: 'P's mother is married to J's husband' implies J's husband is P's father; however, it does not tell whether J is the same person as P's mother (monogamy) or a different spouse (making J a stepmother). Children counts do not disambiguate identities.Concept / Approach:Enumerate models consistent with both statements and observe whether P→J is unique (mother vs stepmother vs other).Step-by-Step Solution:Model A: J = P's mother (monogamous couple). Then P is J's child.Model B: J ≠ P's mother (another spouse of P's father). Then J is P's stepmother.Both models satisfy the statements and are non-contradictory.Verification / Alternative check:Sibling info (I) is compatible with either model; the 'one son and two daughters' clause does not resolve J's identity.Why Other Options Are Wrong:A/B/C/E: No single statement or their combination pins a unique relationship.Common Pitfalls:Assuming spouses are unique without explicit mention; ignoring step-relations.Final Answer:D — Together not sufficient.

Question 16

Rows — In a row of 50 students, how many are between Suresh and Mohan? Statements: I. Suresh is 12th from the left end; Mohan is 17th from the right end. II. Suresh is six places away from Jayesh, who is 20th from the left end.

Accepted Answer

Introduction / Context:We must compute the number of students strictly between two given positions in a fixed-size row.Given Data / Assumptions:Total = 50 students.I: Suresh = 12th from left; Mohan = 17th from right.II: Suresh is 6 away from Jayesh (20th from left).Concept / Approach:Convert Mohan's right-based rank to a left-based rank; then apply the between formula: |p−q| − 1.Step-by-Step Solution:From I: Mohan's left-rank = 50 − 17 + 1 = 34.Between count = 34 − 12 − 1 = 21.From II alone: Suresh's position is either 14th or 26th; without Mohan's absolute rank, the between count is indeterminate.Verification / Alternative check:Positions are valid within bounds [1,50].Why Other Options Are Wrong:B/C/D/E: II lacks Mohan’s absolute position; I already yields a unique number.Common Pitfalls:Forgetting the +1 when converting right-to-left rank; forgetting to subtract 1 for exclusivity.Final Answer:A — Statement I alone suffices (21 students between them).

Question 17

Directions — Village P is in which direction with respect to village R? Statements: I. R is to the south-east of T; T is to the north of P. II. Q is to the south of P and to the south-west of R.

Accepted Answer

Introduction / Context:Determine the bearing of P relative to R using qualitative compass relations. Distances are unspecified.Given Data / Assumptions:I: R is SE of T; T is north of P.II: Q is south of P and SW of R (so R is NE of Q).No distances or collinearity are given.Concept / Approach:Place T as a reference and explore feasible coordinates; assess whether P→R is uniquely determined.Step-by-Step Solution:From I: With T at (0,0), P lies at (0, −a). R lies at (+b, −b). The vector from R to P is (−b, −a + b); its quadrant depends on the relative sizes of a and b, so the bearing is ambiguous.From II: R is NE of Q; P is north of Q. This still does not fix a and b; many consistent layouts exist.Combining I and II does not relate a and b numerically; P could be west, south-west, or south of R depending on magnitudes.Verification / Alternative check:Construct two coordinate examples that satisfy both statements but yield different P→R directions; ambiguity persists.Why Other Options Are Wrong:A/B/C/E: Insufficient constraints to fix a unique bearing.Common Pitfalls:Assuming equal distances without mention; forcing right triangles not stated.Final Answer:D — Together not sufficient.

Question 18

Heights — Among A, B, C, D, E (all different heights), who is the third tallest? Statements: I. E is shorter than only B. II. C is taller than only A.

Accepted Answer

Introduction / Context:We need the 3rd tallest given two partial order constraints among five distinct heights.Given Data / Assumptions:I: E is below only B ⇒ B is tallest (1st), E is 2nd.II: C is above only A ⇒ A is 5th (shortest), C is 4th.Concept / Approach:Merge the two chains to place the remaining person (D).Step-by-Step Solution:From I alone: 1st=B, 2nd=E; 3rd could be C or D or A (insufficient).From II alone: 4th=C, 5th=A; 1st–3rd unresolved (insufficient).Combining: B (1st), E (2nd), C (4th), A (5th). The only remaining slot (3rd) must be D ⇒ unique.Verification / Alternative check:Distinctness is respected; no conflicts arise.Why Other Options Are Wrong:A/B/C: Either statement alone does not fix the 3rd position.E: Although the combination is sufficient, the wording 'necessary' highlights that neither alone suffices; the classification corresponds to needing both.Common Pitfalls:Misplacing C above E; ignoring that II fixes bottom two strictly.Final Answer:D — Both statements together are necessary (3rd tallest is D).

Question 19

Siblings — How many sisters does K have? Statements: I. M is a sister of K. II. K’s mother has three children.

Accepted Answer

Introduction / Context:We need the exact count of K's sisters from family-size and one-known-sister facts.Given Data / Assumptions:I: K has at least one sister (M).II: Total children of K’s mother = 3 (includes K).Concept / Approach:Enumerate possibilities for the third child’s gender.Step-by-Step Solution:From I alone: #sisters ≥ 1, but could be 1 or 2; insufficient.From II alone: With 3 children total, the composition is unknown; #sisters for K could be 0, 1, or 2; insufficient.Together: We know one sister (M) and one additional sibling whose gender is unknown. Therefore, #sisters ∈ {1,2}. Not unique.Verification / Alternative check:Case A: (K male), siblings = {M (F), X (M)} → 1 sister. Case B: (K male), siblings = {M (F), X (F)} → 2 sisters. Both obey I & II.Why Other Options Are Wrong:A/B/C/E: Neither statement alone suffices; even together they do not fix a unique number.Common Pitfalls:Assuming genders not stated; presuming K’s own gender.Final Answer:D — Together not sufficient.

Question 20

Rows — In a row of 30 students facing North, what is R's position from the left end? Statements: I. There are 12 students between R and Q. II. T is 10th from the right end; there are 16 students between T and R.

Accepted Answer

Introduction / Context:Compute an absolute position using a fixed-size row and separations.Given Data / Assumptions:Total = 30 students.I: Only distance between R and Q (no anchors).II: T is 10th from right; 16 between T and R.Concept / Approach:Convert T's right-rank to a left-rank, then apply the 'k between' offset (k+1) and validate side feasibility.Step-by-Step Solution:From II: T's left-rank = 30 − 10 + 1 = 21.If R is to the right of T: R = 21 + 16 + 1 = 38 (invalid > 30). Hence R must be to the left: R = 21 − 16 − 1 = 4.Thus R is 4th from the left. Statement I alone has no absolute anchor and is insufficient.Verification / Alternative check:Bound checks ensure only the left-side placement is feasible, giving a unique result.Why Other Options Are Wrong:A/C/D/E: II alone already yields a unique position.Common Pitfalls:Forgetting to test both sides; forgetting the +1 offset.Final Answer:B — Statement II alone suffices (R = 4th from left).

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