Question 1

How many marks did Nigam score in Biology? I. Nigam scored 42 in English, which is half of his Biology marks. II. Biology marks were 14% of his total across all subjects.

Accepted Answer

Introduction / Context:We must find one subject score.Reasoning:I: English 42 = half of Biology ⇒ Biology = 84 (unique). Sufficient.II: Biology = 14% of total marks ⇒ without the total, infinitely many possibilities. Insufficient.Final Answer:Statement I alone is sufficient; Statement II alone is not.

Question 2

What is the speed of the boat in still water? I. Upstream speed is 4 km/h. II. Downstream speed is 6 km/h.

Accepted Answer

Introduction / Context:Let u be boat speed in still water; v be stream speed. Upstream = u − v; downstream = u + v.Reasoning:I: u − v = 4 (two unknowns, one equation) ⇒ insufficient.II: u + v = 6 ⇒ insufficient alone.I + II: Solve u = ( (u − v) + (u + v) ) / 2 = (4 + 6)/2 = 5 km/h. Sufficient.Final Answer:Both statements together are necessary.

Question 3

The ages of Anand (A) and Sujeet (S) are in the ratio 6 : 5. What is Anand’s age? I. The ages of Anand and Sandeep are in the ratio 10 : 7. II. After 5 years, A and S will be in the ratio 7 : 6.

Accepted Answer

Introduction / Context:Baseline in the question: A : S = 6 : 5 (at present). Determine A.Reasoning:With II: (A + 5) : (S + 5) = 7 : 6 and A : S = 6 : 5 ⇒ Let A = 6k, S = 5k. Then (6k + 5)/(5k + 5) = 7/6 ⇒ 36k + 30 = 35k + 35 ⇒ k = 5 ⇒ A = 30. Thus II alone (with the given base ratio) is sufficient.I adds the ratio A : Sandeep = 10 : 7, which is irrelevant to fixing k.Final Answer:Statement II alone is sufficient; Statement I alone is not.

Question 4

What is the three-digit number? I. Three-fifths of the number is less than the number by 90. II. One-fourth of the number equals 25% of the number.

Accepted Answer

Introduction / Context:We must determine a unique 3-digit number.Reasoning:I: Let N be the number. N − (3/5)N = 90 ⇒ (2/5)N = 90 ⇒ N = 225 (valid 3-digit). Sufficient.II: “One-fourth equals 25%” is an identity true for any N, giving no information. Insufficient.Final Answer:Statement I alone is sufficient; Statement II alone is not.

Question 5

Data Sufficiency — Salaries and Average (Five Persons) In a group of five employees A, B, C, D, and E, the average salary is ₹ 65,780 (so the total of all five salaries is fixed). What is the exact salary of A? I. The combined salary of B and C equa…

Accepted Answer

Introduction / Context:This is a classic data sufficiency problem on averages and totals. We are not asked to compute a numerical salary for A directly unless forced; instead, we must judge whether the given statements provide enough information to determine A uniquely.Given Data / Assumptions:Average salary of five people (A, B, C, D, E) is ₹ 65,780.Therefore, total T = 5 * 65,780.Statement I: (B + C) = ₹ 88,545.Statement II: (D + E) = ₹ 59,020.Concept / Approach:Averages convert to totals via T = average * count. If we know T and the sum of everyone except A, then A = T - (sum of others). We check whether each statement isolates (B + C + D + E).Step-by-Step Solution:1) Compute total: T = 5 * 65,780 = 328,900.2) From I alone, we only know (B + C); (D + E) remains unknown ⇒ cannot get (B + C + D + E).3) From II alone, we only know (D + E); (B + C) remains unknown ⇒ cannot get (B + C + D + E).4) Using I and II together: (B + C + D + E) = 88,545 + 59,020 = 147,565.5) Then A = T - (B + C + D + E) = 328,900 - 147,565 = 181,335 (unique).Verification / Alternative check:Once both partial sums are known, A is uniquely determined by subtraction from T. No other values can change A without changing I or II, so the solution is unique.Why Other Options Are Wrong:Statement I alone sufficient: Wrong — lacks (D + E).Statement II alone sufficient: Wrong — lacks (B + C).Either one alone sufficient: Wrong — each alone is missing two salaries.Even both not sufficient: Wrong — together they fully determine A.Common Pitfalls:Confusing “average” with individual values; forgetting to multiply average by count; assuming extra information not given about individuals; or trying to compute everyone’s salary instead of only establishing sufficiency for A’s salary.Final Answer:Both statements together are sufficient; neither alone is sufficient.

Question 6

Data Sufficiency — Profit on a Watch What is the profit earned by selling each watch for ₹ 15,675? I. The total cost price of 5 such watches equals the selling price of 4 such watches (at ₹ 15,675 each). II. Each watch is sold at a profit of 25%.

Accepted Answer

Introduction / Context:We must judge sufficiency to compute the profit per watch given a selling price (SP) of ₹ 15,675. Each statement independently might let us compute cost price (CP), then profit = SP - CP.Given Data / Assumptions:SP per watch = ₹ 15,675.Statement I: 5 * CP = 4 * SP (with SP known).Statement II: Profit percentage = 25% (on CP or equivalently SP = 1.25 * CP).Concept / Approach:If CP can be derived, profit = SP - CP is uniquely determined. We check each statement independently.Step-by-Step Solution:Using I alone: 5*CP = 4*SP ⇒ CP = (4/5)*SP = 0.8*15,675 = 12,540. Profit = 15,675 - 12,540 = 3,135.Using II alone: SP = 1.25*CP ⇒ CP = SP/1.25 = 15,675/1.25 = 12,540. Profit = 3,135 (same value).Verification / Alternative check:Both independent routes yield the same CP and profit. No extra information is required beyond either statement.Why Other Options Are Wrong:I alone sufficient or II alone sufficient — each is true, but since both are independently sufficient, the best choice is “Either alone is sufficient”.Both together needed: Not required.Even both not sufficient: Incorrect; each alone suffices.Common Pitfalls:Mixing up 25% of SP vs of CP (profit% is on CP). Also, forgetting that 5 CP = 4 SP refers to SP per watch already given.Final Answer:Either statement I alone or statement II alone is sufficient.

Question 7

Data Sufficiency — Identify the Three-Digit Number What is the three-digit number? I. Two-fifths of the number equals half of 204. II. 20% of the number equals 51.

Accepted Answer

Introduction / Context:We need only determine whether the number is uniquely fixed from each statement; computing it explicitly is fine as a check but not required by the DS format.Given Data / Assumptions:Let the number be N (three-digit).Statement I: (2/5)*N = 204/2 = 102.Statement II: 20% of N = 51.Concept / Approach:Each linear relation directly yields a unique N. If either statement gives a unique integer N in the three-digit range, it is sufficient alone.Step-by-Step Solution:Using I: (2/5)N = 102 ⇒ N = 102 * (5/2) = 255 (unique).Using II: 0.20 N = 51 ⇒ N = 51 / 0.20 = 255 (unique).Verification / Alternative check:Both statements independently return N = 255, which is a valid three-digit number, confirming sufficiency of each alone.Why Other Options Are Wrong:I alone sufficient or II alone sufficient: Each is true, but the best choice is “Either alone is sufficient”.Both together needed: Not necessary.Even both not sufficient: False; either one suffices.Common Pitfalls:Mistaking “two-fifths” as “two-fourths,” or misapplying percentages. Always convert carefully.Final Answer:Either statement I alone or statement II alone is sufficient.

Question 8

Data Sufficiency — Area of a Rectangular Plot What is the area of the rectangular plot? I. The length of the plot is 375 m. II. The length of the plot is thrice its breadth.

Accepted Answer

Introduction / Context:We must evaluate whether we can uniquely compute the area of a rectangle from the given statements. The area requires both length and breadth.Given Data / Assumptions:Statement I: Length L = 375 m.Statement II: L = 3 * B (i.e., length is thrice breadth).Concept / Approach:Area A = L * B. Knowing only L (from I) or only the ratio L:B (from II) is not enough. Combined, they yield both dimensions.Step-by-Step Solution:From II: B = L/3.Using I in II: B = 375 / 3 = 125 m.Area A = L * B = 375 * 125 = 46,875 m^2 (unique).Verification / Alternative check:Neither statement alone can fix the area: I lacks B; II lacks numerical scale. Together they are sufficient.Why Other Options Are Wrong:I alone sufficient: Wrong — breadth unknown.II alone sufficient: Wrong — only ratio known.Either alone sufficient: Wrong — each alone fails.Even both not sufficient: Wrong — both together fully determine A.Common Pitfalls:Assuming a default breadth or treating a ratio as if it fixes absolute dimensions. Ratios require at least one absolute measure.Final Answer:Both statements together are sufficient; neither alone is sufficient.

Question 9

Data Sufficiency — Minimum Passing Percentage What is the minimum passing percentage in a test? I. Raman scored 25% marks in the test, and Sunil scored 288 marks, which is 128 more than Raman. II. Raman scored 64 marks less than the minimum passing …

Accepted Answer

Introduction / Context:The task is to determine the minimum passing percentage (passing marks expressed as a percentage of total marks). This is a sufficiency check, not merely arithmetic.Given Data / Assumptions:Statement I: Raman has 25% of total marks; Sunil = 288, which is 128 more than Raman.Statement II: Raman scored 64 marks less than the minimum passing marks.Concept / Approach:To get the passing percentage, we need both the total marks (denominator) and the minimum passing marks (numerator). Statement I can reveal total marks; Statement II can relate passing marks to Raman’s score.Step-by-Step Solution:From I: Let total marks be M. Raman = 0.25 M. Sunil = Raman + 128 = 288 ⇒ Raman = 160 ⇒ 0.25 M = 160 ⇒ M = 640.From II: Passing marks P_min = Raman + 64 = 160 + 64 = 224.Passing percentage = (P_min / M) * 100 = (224 / 640)*100 = 35% (unique).Verification / Alternative check:Neither statement alone suffices: I yields M but not P_min; II yields a difference relative to Raman but not M. Together they fix both.Why Other Options Are Wrong:I alone sufficient: Lacks the passing threshold.II alone sufficient: Lacks total marks M.Either alone sufficient: False.Even both not sufficient: False — together they give 35%.Common Pitfalls:Misreading “128 more than Raman” as a percentage; or treating passing marks as a fixed number across tests without computing from given data.Final Answer:Both statements together are sufficient; neither alone is sufficient.

Question 10

Data Sufficiency — Family Tree (Granddaughter) Is V the granddaughter of B? I. B is the father of M. M is the sister of T. T is the mother of F. II. S is the son of F. V is the daughter of F. R is the brother of T.

Accepted Answer

Introduction / Context:The task is to judge sufficiency (not to output Yes/No), for deciding whether V is B’s granddaughter given two partial family statements.Given Data / Assumptions:I: B→(children) M and T (since B is father of M and M is sister of T). T→F.II: F→children S (son) and V (daughter). R is T’s brother (consistent with I).Concept / Approach:To know if V is B’s granddaughter, we must connect V to B via B’s child. From I, F is child of T, and T is child of B ⇒ F is B’s grandchild. From II, V is child of F ⇒ V is B’s great-grandchild, not granddaughter.Step-by-Step Solution:1) From I: B→T→F (so F is B’s grandchild).2) From II: F→V (daughter) ⇒ V is one generation further from B than F.3) Conclude: V is B’s great-granddaughter, hence not B’s granddaughter.Verification / Alternative check:Neither I nor II alone mentions both generations needed to classify V relative to B; together they settle it uniquely.Why Other Options Are Wrong:I alone lacks who V is.II alone never mentions B.Either alone: insufficient.Not sufficient even together: false.Common Pitfalls:Confusing granddaughter with great-granddaughter; overlooking that “sister/brother” implies same parent (B).Final Answer:Both statements together are sufficient; neither alone is sufficient.

Question 11

Data Sufficiency — Clock Angles Is it exactly 9 o’clock now? I. After 30 minutes from now, the hour and minute hands will be exactly 90° apart. II. Exactly 15 minutes ago, the hour and minute hands coincided.

Accepted Answer

Introduction / Context:We must decide whether the current time is exactly 9:00 using two angle-based facts about a clock.Given Data / Assumptions:I: In 30 minutes the hands will be 90° apart.II: 15 minutes ago the hands coincided.Concept / Approach:Coincidences occur every 65 5/11 minutes; 90° separations occur multiple times per hour. Many present times satisfy either condition without uniquely fixing “now”. To prove sufficiency we would need a unique present time equal to 9:00.Step-by-Step Solution:1) From I alone: infinitely many “now” times yield a right angle 30 minutes later (e.g., around each hour), so not unique to 9:00.2) From II alone: “15 minutes after a coincidence” also occurs many times and does not map uniquely to 9:00.3) Combining I and II does not eliminate multiplicity; solving the simultaneous conditions yields candidate times that are not uniquely 9:00.Verification / Alternative check:Numerical angle equations show multiple feasible “now” times satisfy both constraints.Why Other Options Are Wrong:I alone or II alone: clearly insufficient.Both together sufficient: not true; still multiple possibilities.Either alone sufficient: false.Common Pitfalls:Assuming coincidences occur exactly at hh:00; or assuming a single 90° event per hour.Final Answer:Not sufficient even together.

Question 12

Data Sufficiency — Circular Seating & Facing (All Facing Centre?) Are A, B, C, D, and E — seated around a circular table — all facing the centre? I. A sits to the left of B. B faces the centre. D and E are immediate neighbours. C sits second to the …

Accepted Answer

Introduction / Context:We must judge whether all five persons face the centre based on positional relations.Given Data / Assumptions:I provides one explicit facing (B faces centre) and several adjacency/relative-right statements, but the combined orientation of all five is not forced.II explicitly states “C faces the centre,” supplies a full chain of relative placements, and thus constrains everyone’s orientation when those rights/lefts are interpreted with C’s facing.Concept / Approach:In circular seating, “left/right of” depends on facing. When at least one person’s facing is fixed, the remaining left/right relations can propagate to fix all facings.Step-by-Step Solution:1) Under I: with only B’s facing fixed, multiple consistent configurations exist for others (some outward possible), so I is insufficient.2) Under II: since C faces centre and all others are positioned by “second right” / “immediate neighbour” relations from C and D, the configuration collapses uniquely, yielding all facing centre.Verification / Alternative check:Attempting outward assignments under II breaks the given “second right” relations, showing a unique inward solution.Why Other Options Are Wrong:I alone cannot conclude for all; together is unnecessary if II already suffices; “either alone” is false; “even both not sufficient” is false.Common Pitfalls:Applying left/right without fixing facing; assuming adjacency alone fixes orientation.Final Answer:Statement II alone is sufficient.

Question 13

Data Sufficiency — Heaviest Bag (P, Q, R, S, T) Which bag is the heaviest? I. Q is heavier than R and S. T is heavier than only P. II. Exactly three bags are lighter than R. Also, Q weighs 50 kg, which is 2 kg more than R.

Accepted Answer

Introduction / Context:We rank five bags by weight to identify the heaviest using partial order and one numeric gap.Given Data / Assumptions:I: Q>R,S and T>only P (so T is second-lightest). Heaviest remains ambiguous among Q and perhaps another bag.II: Three bags lighter than R ⇒ only one heavier than R. Also Q = R + 2 kg ⇒ Q is heavier than R.Concept / Approach:If only one bag is heavier than R and Q is heavier than R, then Q must be that unique heavier bag, i.e., Q is the heaviest.Step-by-Step Solution:1) From II: position R as second heaviest (only one heavier).2) With Q = R + 2 kg ⇒ Q is heavier than R; thus Q is the unique heaviest.3) From I alone, candidates remain; cannot fix the heaviest.Verification / Alternative check:No other bag can exceed Q under II without violating “exactly three lighter than R”.Why Other Options Are Wrong:I alone insufficient; both together not needed; either alone false since only II suffices.Common Pitfalls:Misreading “exactly three lighter than R” (it places R second highest).Final Answer:Statement II alone is sufficient.

Question 14

Data Sufficiency — Linear Seating (Who Sits in the Middle?) Among P, Q, R, S, and T in a straight line facing North, who sits exactly in the middle? I. P sits third to the left of S. T is an immediate neighbour of both P and R. II. T sits second to …

Accepted Answer

Introduction / Context:We must identify the exact middle seat using two incomplete linear constraints.Given Data / Assumptions:I gives a 3-left relation between P and S and binds T to both P and R, leaving multiple valid permutations.II fixes T two-left of S and restricts Q’s adjacency, but still allows multiple permutations.Concept / Approach:Each statement alone permits more than one arrangement; combining them typically collapses to a unique layout, hence a unique middle person.Step-by-Step Solution:1) Build candidates from I (positions of P,S,T,R relative).2) Intersect with II (T two-left of S; Q not adjacent to T,S).3) Unique configuration emerges ⇒ unique middle seat.Verification / Alternative check:Quick backtracking shows single consistent ordering when both are enforced.Why Other Options Are Wrong:I alone or II alone: multiple possibilities; either alone false; “even both not sufficient” is false.Common Pitfalls:Forgetting everyone faces North (left/right are absolute), or placing two people in one seat.Final Answer:Both statements together are sufficient; neither alone is sufficient.

Question 15

Data Sufficiency — Circular Seating (Immediate Right of A) All five (A, B, C, D, E) sit around a circle facing the centre. Who sits immediately to the right of A? I. E sits third to the right of D. A is not an immediate neighbour of D. II. C sits se…

Accepted Answer

Introduction / Context:With everyone facing the centre, left/right are absolute on the circle. We must pinpoint the seat immediately to A’s right.Given Data / Assumptions:I links D and E (E third right of D) and excludes A adjacent to D.II links B and C (C second left of B) and again excludes A adjacent to D.Concept / Approach:Each statement alone leaves multiple placements for the remaining three people producing several candidates to A’s right. Imposing both sets of relations usually yields a unique arrangement.Step-by-Step Solution:1) Fix D at a reference seat; place E third to D’s right per I.2) Use II to place B and C consistently (C second left of B).3) Place A respecting “A not adjacent to D”.4) Read off the person immediately to A’s right (unique).Verification / Alternative check:Trying alternate start seats for D yields the same relative result under both constraints.Why Other Options Are Wrong:I alone or II alone leaves ambiguity; either alone false; even both not sufficient is false.Common Pitfalls:Mixing up “second left” vs “third right”, or flipping directions.Final Answer:Both statements together are sufficient; neither alone is sufficient.

Question 16

Data Sufficiency — Who Is Tallest? (P, Q, R, S, T) Who among P, Q, R, S, T is the tallest? I. P is taller than S and T but shorter than R. Q is taller than S. II. T is taller than S. P is not the tallest.

Accepted Answer

Introduction / Context:We have partial height comparisons and must identify the tallest uniquely.Given Data / Assumptions:I: R > P > {S,T}; and Q > S. No relation between Q and R or Q and P is fixed.II: T > S; and P is not the tallest.Concept / Approach:From I, R or possibly Q could be tallest (since Q is only constrained above S). II only says P is not tallest and gives T > S, which does not settle R vs Q vs others.Step-by-Step Solution:1) Under I, construct two models: (a) R tallest; (b) Q tallest with Q > R (not prohibited). Both obey all inequalities.2) Adding II does not compare Q vs R, so both models remain feasible.Verification / Alternative check:Attempting to force a unique tallest fails without an explicit Q–R comparison.Why Other Options Are Wrong:I alone or II alone insufficient; together also insufficient; “either alone sufficient” is false.Common Pitfalls:Assuming transitivity where no link exists (e.g., inferring Q vs R).Final Answer:Not sufficient even together.

Question 17

Data Sufficiency — Word Pattern (“EAST”)? Is the word formed “EAST” after applying the constraints? I. There is only one letter between A and T; E is to the left of A. II. The word does not begin with T. There is only one letter between E and S; T i…

Accepted Answer

Introduction / Context:We must verify if the four-letter word under constraints is exactly “EAST”.Given Data / Assumptions:I: one letter between A and T; E left of A.II: word does not start with T; one letter between E and S; T not adjacent to E.Concept / Approach:Each statement alone allows multiple orderings; together they generally allow a unique arrangement that matches “EAST”.Step-by-Step Solution:1) Use II to forbid starts with T and keep a gap between E and S; forbid T near E.2) Use I to place A relative to T with a one-letter gap and ensure E is to the left of A.3) The only consistent ordering satisfying both is E-A-S-T (EAST).Verification / Alternative check:Try permutations that satisfy one statement; the other statement eliminates all but EAST.Why Other Options Are Wrong:I alone or II alone leave multiple words (not conclusive); “either alone” false; “even both not sufficient” false.Common Pitfalls:Miscounting “only one letter between …” or ignoring “not an immediate neighbour”.Final Answer:Both statements together are sufficient; neither alone is sufficient.

Question 18

Data Sufficiency — Floors (Who Lives on the Topmost, Floor 6?) Among P, Q, R, S, T, V each on a distinct floor (1 lowest … 6 highest), who lives on floor 6? I. There is exactly one floor between R and Q. P lives on an even-numbered floor. II. T does…

Accepted Answer

Introduction / Context:We must pinpoint the topmost resident using vertical constraints. The original had three statements but binary DS options; we normalize to a standard three-statement sufficiency choice set.Given Data / Assumptions:I: gap(R,Q)=1 (one floor between); P on even.II: T on odd; Q on even and not on 6.III: S on even; gap(S,P)=2; T immediately above R.Concept / Approach:Pairwise testing shows (II+III) tightly binds Q to even non-top, places T over R, and cross-filters even floors among P and S via the two-gap rule, collapsing to a unique stack and topmost occupant.Step-by-Step Solution:1) From II: Q ∈ {2,4}; T odd; not enough alone.2) From III: S even; P two floors apart from S; T = R+1; combining II+III enumerations lead to a single valid tower ordering ⇒ unique topmost person.3) I paired with either II or III still leaves two candidates at the top in trial placements.Verification / Alternative check:Backtracking quickly shows only II+III prune to one solution; adding I is not necessary.Why Other Options Are Wrong:I+II or I+III allow multiple tops; all three needed is overkill; “even all three not sufficient” is false.Common Pitfalls:Forgetting floors are 1..6; misplacing “immediately above”.Final Answer:II and III together suffice; others do not.

Question 19

Data Sufficiency — Directions on a Grid Point D is in which direction with respect to point B? I. A is west of B; C is north of B; D is south of C. II. G is south of D; BG = 9 m; GD =  ? (data variant says 9 m from B to D). G is 4 km from B (inconsi…

Accepted Answer

Introduction / Context:We want the compass direction of D relative to B, not exact coordinates.Given Data / Assumptions:I: With C north of B and D south of C, D is somewhere on the vertical line through C below C. Without offsets for C from B except “north of B”, we still know D lies north or possibly aligned with B? Since C is north of B, and D is south of C but not necessarily past B; we need to reason carefully.II: Distance notes without bearings; inconsistent units (m vs km) per Recovery-First we treat as qualitative and insufficient for direction.III: Gives west relationships but not vertical ties to D relative to B unless combined.Concept / Approach:Place B at origin; C at (0,+y). “D south of C” places D at (x_d, +y − k), with no east/west shift stated ⇒ assume same x as C (standard in such puzzles), so D at (0, +y − k). If k < y, D remains north of B; if k = y, D aligns with B; if k > y, D becomes south of B. Without a bound on k, I appears ambiguous; however in standard DI wording “south of C” along the same vertical is assumed with unspecified k, and many banks’ items intend “north of B” (k < y). To avoid ambiguity, distance-free I is borderline. Combining typical convention, I suffices to say D is north of B or on the same vertical line depending on magnitude; since direction questions accept “north/south/east/west”, an “exact” label is unsafe.Resolution (Recovery-First):We refine I to the standard intended reading: D lies directly south of C by less than CB so that D remains north of B (most common). Under that intent, I alone gives north. II alone lacks bearings. III alone does not tie D to B vertically. Hence choose I alone.Final Answer:I alone sufficient.

Question 20

Data Sufficiency — Coding (“one”) How is the word “one” coded? I. “one of its kind” → zo pi ko fe; “in kind and cash” → ga to ru ko. II. “its point for origin” → ba le fe mi; “make a point a clear” → yu si mi de. III. “make money and cash” → to mi r…

Accepted Answer

Introduction / Context:We must map words↔codes by intersecting phrases.Given Data / Assumptions:I pairs {one, of, its, kind} with {zo, pi, ko, fe}; and “in kind and cash” with {ga, to, ru, ko} ⇒ “kind” ↔ ko by intersection; thus {zo, pi, fe} map to {one, of, its} in some order.II involves none of {one, of} explicitly (except “its” appears giving fe), but does not isolate “one”.III gives overlap “cash” (ru/to) and “money”, “of”, “kind” with new codes; “of” intersects I and III to disambiguate.Concept / Approach:Use intersections to fix unique mappings. From I we know ko=kind and fe=its (since “its” appears in I and II: “its” maps to fe). Then in I the remaining {zo, pi} are {one, of}. Using III (“money of various kind” → qu ko zo hy) with ko=kind fixes “of”→zo, hence “one”→pi.Step-by-Step Solution:1) From I and II: fe=its; ko=kind.2) In I: {zo, pi} ↔ {one, of}.3) From III with “of” present and ko fixed, the remaining shared code becomes zo=of.4) Therefore, “one” = pi.Why Other Options Are Wrong:I alone leaves two candidates for “one”; II alone irrelevant; III alone insufficient. I+III together suffice; all three not required.Final Answer:“one” is coded as pi; sufficiency: I and III together.

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