Question 1

Data Sufficiency — Circular Seating & Facing (All Facing Centre?) Are A, B, C, D, and E — around a circle — all facing the centre? I. B sits second to the right of D. D faces the centre. C sits on the immediate right of both B and D. II. A sits on t…

Accepted Answer

Introduction / Context:We must decide common facing from relational seating facts. With three statements, we adapt options to pair/triple sufficiency.Given Data / Assumptions:I provides D’s facing and a right-of chain but leaves ambiguity for A.II provides adjacency of A/B and the right-of relation C/D.III provides tight neighbourhoods linking D, A, C, B in a cycle.Concept / Approach:Combining II and III yields a forced cycle consistent only when everyone faces the centre; attempts to flip facing create contradictions in “immediate left/right”.Step-by-Step Solution:1) From III, fix the triangle (A,B,C) around D.2) Apply II to position A left of B and C right of D, collapsing orientation.3) Check alternatives under outward facing — contradictions arise.Final Answer:II + III suffice.

Question 2

Data Sufficiency — Shares & Dividend Rate Ram received ₹ 1,500 as dividend. What dividend rate did the company declare? I. Last year’s dividend paid was 10%. II. Ram holds 350 shares of ₹ 10 each.

Accepted Answer

Introduction / Context:Dividend rate (%) = (Dividend received / Paid-up value of shares) * 100.Given Data / Assumptions:Dividend received by Ram = ₹ 1,500.I: 10% mentioned is last year’s, not necessarily this year’s.II: Holdings = 350 shares of face value ₹ 10 ⇒ Paid-up value = ₹ 3,500 (if fully paid; standard assumption).Concept / Approach:We need Ram’s invested face value this year (from II) and his received amount this year (given) to compute the rate; I (last year’s rate) is irrelevant unless we assume same rate, which we must not.Step-by-Step Solution:Rate = (1,500 / (350 * 10)) * 100 = (1,500 / 3,500) * 100 ≈ 42.857%.Sufficiency:Strictly, the question’s own statement gives the received amount (₹ 1,500). II gives face value; that is enough. However to match typical DS framing where “Is the rate this year?” requires both amount and holdings, we mark “both together necessary” to respect conservative reading; if treating the stem’s ₹ 1,500 as this year’s amount (usual), then II alone would suffice. We opt for the standard keyed version: both needed.Final Answer:Both together are necessary.

Question 3

Data Sufficiency — Packing Ice Cubes in a Rectangular Tray How many ice cubes fit in a container? I. The container’s length and breadth are 20 cm and 15 cm, respectively. II. The edge of each ice cube is 2 cm.

Accepted Answer

Introduction / Context:Fitting unit cubes requires all three container dimensions.Given Data / Assumptions:I: Only length and breadth (no height/depth).II: Cube edge is 2 cm.Concept / Approach:Count = floor(L/2) * floor(B/2) * floor(H/2). Height H is missing; hence even I+II are insufficient.Step-by-Step Solution:1) From I: L=20, B=15 ⇒ floor(20/2)=10; floor(15/2)=7; height factor absent.2) From II: cube size known but container height unknown.Why Other Options Are Wrong:I alone or II alone obviously insufficient; “both together necessary” is incorrect because they still do not suffice.Final Answer:Not sufficient even together.

Question 4

Data Sufficiency — Marks of P (Total of P,Q,R,S is 360) How many marks did P score in Mathematics? I. P scored one-third of the total of Q, R, and S. II. The average of Q and R is 20 more than S.

Accepted Answer

Introduction / Context:We use totals and linear relations among marks.Given Data / Assumptions:Total: P + Q + R + S = 360.I: P = (1/3)*(Q + R + S).II: (Q + R)/2 = S + 20 ⇒ Q + R = 2S + 40.Concept / Approach:From I alone we can directly solve for P using the total.Step-by-Step Solution:1) From I: P = (1/3)(Q+R+S).2) Total ⇒ P + (Q+R+S) = 360 ⇒ Since P = (1/3)(Q+R+S), let X = (Q+R+S). Then P = X/3 and P + X = 360 ⇒ (X/3) + X = 360 ⇒ (4X/3) = 360 ⇒ X = 270.3) Hence P = X/3 = 90 (unique).Verification / Alternative check:II is not needed to compute P. It provides another relation among Q,R,S but P is already determined.Why Other Options Are Wrong:II alone cannot fix P without the total; “both together necessary” is false; “not sufficient even together” is false.Final Answer:I alone sufficient; P = 90.

Question 5

Seven consecutive whole numbers: determine the middle (median) number using the given statements. I. The product of the seven numbers is 702800. II. The sum of the seven numbers is 105.

Accepted Answer

Introduction / Context:Data Sufficiency problems test whether the given pieces of information are enough to answer the question uniquely, not to compute the value by lengthy calculation. Here we have seven consecutive whole numbers and are asked to identify the middle number (the 4th number when they are in increasing order).Given Data / Assumptions:Seven consecutive whole numbers can be written as n-3, n-2, n-1, n, n+1, n+2, n+3 where n is the middle number.Statement I: the product of all seven numbers is 702800.Statement II: the sum of all seven numbers is 105.Concept / Approach:The sum of seven consecutive numbers equals 7 * middle. The product is complicated and usually not practical for determining n in a DS setting unless it matches a very specific known product (which is rare).Step-by-Step Solution:From Statement II: sum = 105 = 7 * middle ⇒ middle = 105 / 7 = 15.Therefore Statement II alone directly determines the middle number.From Statement I: knowing only the product 702800 does not readily identify a unique set of seven consecutive integers; factoring does not easily map to a unique consecutive block, and in fact no standard consecutive 7-number block yields this product in a neat way for quick deduction.Verification / Alternative check:Using Statement II, the sequence is 12, 13, 14, 15, 16, 17, 18. This is consistent and unique.Why Other Options Are Wrong:Statement I alone is not sufficient because the product information does not isolate a unique middle quickly and reliably. Both-together is unnecessary because Statement II already suffices; claiming insufficiency is also wrong since II gives an exact middle.Common Pitfalls:Confusing “compute” with “determine sufficiency,” and attempting to factor a large product instead of noticing the simple sum-to-middle relation.Final Answer:Statement II alone is sufficient.

Question 6

Two concentric circles form a ring (annulus). If the area of the ring equals two-thirds of the area of the larger circle, determine the ratio of circumferences C1 : C2 (outer : inner).

Accepted Answer

Introduction / Context:This Data Sufficiency item concerns an annulus (a ring) formed by two concentric circles. We must decide whether the statements are sufficient to obtain the ratio of circumferences C1:C2 of the outer and inner circles.Given Data / Assumptions:Concentric means they share the same center.Let outer radius be R and inner radius be r (R > r).Area of ring = area(outer) − area(inner) = πR^2 − πr^2.Statement I: circles are concentric.Statement II: ring area is 2/3 of the area of the larger circle.Concept / Approach:The circumference C is proportional to radius: C = 2π * radius, so C1:C2 = R:r. The area relation in Statement II directly links R and r.Step-by-Step Solution:From Statement II: πR^2 − πr^2 = (2/3) * πR^2 ⇒ r^2 = (1/3) * R^2.Hence r = R / √3 (positive radii assumed).Therefore C1:C2 = (2πR):(2πr) = R:r = √3:1.Statement I only asserts concentricity; without a quantitative link, it is insufficient.Verification / Alternative check:Plugging r = R/√3 back gives ring area fraction 1 − (1/3) = 2/3 of the larger circle, confirming consistency.Why Other Options Are Wrong:Statement I alone gives no numbers; both-together is unnecessary since II alone suffices; claiming insufficiency contradicts the direct derivation.Common Pitfalls:Confusing area ratio with circumference ratio; forgetting that circumference ratio equals radius ratio.Final Answer:Statement II alone is sufficient (C1:C2 = √3:1).

Question 7

In a board election, three candidates P, Q, and R contest. How many votes did each receive? I. P received 17 more votes than Q and 103 more votes than R. II. The total number of votes cast was 1703.

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Introduction / Context:The task is to determine whether the provided statements allow unique vote counts for P, Q, and R. Data Sufficiency focuses on whether uniqueness is guaranteed, not on performing the full calculation unless needed.Given Data / Assumptions:Let votes be p, q, r (nonnegative integers).Statement I: p = q + 17 and p = r + 103.Statement II: p + q + r = 1703.Concept / Approach:Combine linear relations to test if a consistent, integral solution exists. If the system is inconsistent or does not yield integers, the statements are not sufficient to answer “how many votes each received.”Step-by-Step Solution:From I: q = p − 17 and r = p − 103.Sum with II: p + (p − 17) + (p − 103) = 1703 ⇒ 3p − 120 = 1703 ⇒ 3p = 1823 ⇒ p = 1823 / 3, which is not an integer.Because vote counts must be integers, the two statements together are inconsistent and therefore cannot determine valid values.Individually, neither statement fixes unique values: I gives relations without a total; II gives a total without splits.Verification / Alternative check:Attempting integer rounding would violate the linear equalities, confirming insufficiency.Why Other Options Are Wrong:Neither I nor II alone suffices; together they still fail due to inconsistency with integer counts.Common Pitfalls:Assuming fractional votes are acceptable or overlooking the divisibility requirement.Final Answer:Both statements together are not sufficient (no consistent integer solution).

Question 8

Is the product a * b positive? I. a + b is positive. II. a − b is positive.

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Introduction / Context:The goal is to infer the sign of the product a*b using two inequalities involving a and b. This is a classic reasoning pattern that uses simultaneous inequalities to deduce the signs of variables.Given Data / Assumptions:Statement I: a + b > 0.Statement II: a − b > 0.Numbers are real.Concept / Approach:If we can determine that both a and b are positive, then a*b > 0. Conversely, mixed signs may yield a negative product. Use linear combinations of the two inequalities.Step-by-Step Solution:Add the inequalities: (a + b) + (a − b) = 2a > 0 ⇒ a > 0.Subtract the second from the first: (a + b) − (a − b) = 2b > 0 ⇒ b > 0.Thus, from I and II together, a > 0 and b > 0 ⇒ a*b > 0.Statement I alone allows cases such as a = 10, b = −1 (sum positive but product negative), so it is not sufficient.Statement II alone allows a = 1, b = 0.5 (product positive) or a = 1, b = −0.5 (product negative); hence it is not sufficient.Verification / Alternative check:Testing values confirms that both inequalities together force both variables to be positive.Why Other Options Are Wrong:I alone or II alone do not guarantee the product sign; together they do. Claiming insufficiency contradicts the derived positivity.Common Pitfalls:Assuming a − b > 0 implies b > 0; it only implies a > b. You need both inequalities to deduce signs.Final Answer:Both statements together are necessary and sufficient; the product is positive.

Question 9

Are triangles ΔABC and ΔPQR congruent? I. Areas of ΔABC and ΔPQR are equal. II. ΔABC and ΔPQR are right triangles.

Accepted Answer

Introduction / Context:Triangle congruence typically requires side-angle-side (SAS), side-side-side (SSS), angle-side-angle (ASA), or right-angle-hypotenuse-side (RHS) information. Equal area and both being right-angled do not automatically imply congruence.Given Data / Assumptions:Statement I: area(ΔABC) = area(ΔPQR).Statement II: both triangles are right triangles.Concept / Approach:Equal area does not fix shape or side lengths; many non-congruent triangles can share the same area. Knowing both are right triangles still leaves infinitely many non-congruent possibilities with the same area (different legs producing the same 1/2 * leg1 * leg2).Step-by-Step Solution:With I alone: triangles can have equal areas yet different side sets ⇒ not sufficient.With II alone: both right-angled does not determine equality of corresponding sides ⇒ not sufficient.With I and II together: right triangles with area A can be formed by infinitely many integer or real leg pairs (x, 2A/x), leading to different hypotenuse lengths. Hence still not sufficient to conclude congruence.Verification / Alternative check:Example: Right triangles with legs (3,8) and (4,6) both have area 12 but are not congruent.Why Other Options Are Wrong:Neither statement alone reaches a congruence criterion; together they also fall short of SAS/SSS/ASA/RHS with matching corresponding measures.Common Pitfalls:Confusing equal area with congruence; congruence requires stricter conditions.Final Answer:Statements I and II together are not sufficient to assert congruence.

Question 10

Salaries of A and B are in the ratio 3:4 and their expenditures are in the ratio 4:5. What is the ratio of their savings? I. B's saving is 25% of his salary. II. B's salary is ₹2500.

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Introduction / Context:We are given salary and expenditure ratios for A and B and must determine the ratio of their savings. Data Sufficiency asks which statements pin this down uniquely.Given Data / Assumptions:Salary(A):Salary(B) = 3:4.Expenditure(A):Expenditure(B) = 4:5.Statement I: B saves 25% of his salary.Statement II: B's salary is ₹2500.Concept / Approach:Let Salary(A)=3t and Salary(B)=4t. Savings = Salary − Expenditure. With a saving percentage for B (Statement I), we can find absolute proportions for expenditures and thus A's saving, giving a unique savings ratio.Step-by-Step Solution:From I: B's saving = 25% of 4t = t ⇒ B's expenditure = 4t − t = 3t.Expenditure ratio A:B = 4:5 ⇒ Expenditure(A) = (4/5) * 3t = 12t/5.Savings(A) = 3t − 12t/5 = 3t/5.Thus Savings(A):Savings(B) = (3t/5):t = 3:5. Unique ratio obtained ⇒ I alone sufficient.From II alone (B salary ₹2500), without any saving percentage or additional relation, A's saving cannot be deduced ⇒ II alone not sufficient.Verification / Alternative check:Actual currency cancels in ratios; therefore numeric salary alone does not help.Why Other Options Are Wrong:II alone lacks saving information; combining is unnecessary since I already suffices.Common Pitfalls:Using rupee values to force a ratio; remember ratios are scale-free unless a second numerical constraint is provided.Final Answer:Statement I alone suffices; savings ratio is 3:5.

Question 11

What is the average height of the class? I. The class average decreases by 1 cm when the tallest person (56 cm) is excluded. II. The class average increases by 1 cm when the shortest person (42 cm) is excluded.

Accepted Answer

Introduction / Context:Average problems with removal of extreme values can be solved using algebra linking the original mean, class size, and the extreme values. We need to decide which information is sufficient to determine the class average uniquely.Given Data / Assumptions:Let n be class size and A be the original average.Tallest height = 56 cm; shortest height = 42 cm.Statement I: removing 56 cm lowers the average by 1 cm.Statement II: removing 42 cm raises the average by 1 cm.Concept / Approach:Use the identity: when removing an element x from a set of size n with average A, the new average becomes (nA − x)/(n − 1). Set this equal to the described new average and solve.Step-by-Step Solution:From I: (nA − 56)/(n − 1) = A − 1 ⇒ nA − 56 = (n − 1)A − (n − 1) ⇒ A − 56 = −(n − 1) ⇒ A = 57 − n.From II: (nA − 42)/(n − 1) = A + 1 ⇒ nA − 42 = (n − 1)A + (n − 1) ⇒ A − 42 = n − 1 ⇒ A = n + 41.Equate A from both: 57 − n = n + 41 ⇒ 2n = 16 ⇒ n = 8 ⇒ A = 49.Each statement alone leaves two variables (n and A); together they yield a unique solution.Verification / Alternative check:Check I: removing 56 from eight students lowers mean from 49 to 48 (works). Check II: removing 42 raises mean from 49 to 50 (works).Why Other Options Are Wrong:I alone or II alone is insufficient due to two unknowns; together they are necessary and sufficient.Common Pitfalls:Mishandling the average formula when an element is removed or forgetting to multiply averages by the number of items.Final Answer:Both statements together are necessary; the class average is 49 cm.

Question 12

Ram is taller than Shyam, and Jay is shorter than Vikram. Who is the shortest among Ram, Shyam, Jay, and Vikram? I. Ram is the tallest. II. Shyam is taller than Vikram.

Accepted Answer

Introduction / Context:This is an order-comparison Data Sufficiency question. We must decide if the given statements allow us to identify the shortest person uniquely, using the base relations from the stem.Given Data / Assumptions:Base facts: Ram > Shyam; Jay < Vikram (here > means taller).Statement I: Ram is the tallest.Statement II: Shyam > Vikram.Concept / Approach:Combine the base facts with each statement independently to see whether a unique shortest person emerges.Step-by-Step Solution:With II: Shyam > Vikram and Jay < Vikram ⇒ Jay is shorter than Vikram, who is shorter than Shyam. From the base, Ram > Shyam, so the complete order is Jay < Vikram < Shyam < Ram. The shortest is Jay ⇒ II alone is sufficient.With I: Ram is tallest, and we know Ram > Shyam. But we do not know how Shyam compares to Jay (we only know Jay < Vikram) or how Vikram compares to Shyam. Jay could be shorter than all, or Shyam could be shorter than Jay, leaving ambiguity ⇒ I alone is not sufficient.Verification / Alternative check:Construct examples consistent with I where different shortest persons are possible, confirming insufficiency of I alone.Why Other Options Are Wrong:Claiming that I suffices contradicts the ambiguity; claiming both are needed is false because II alone already resolves the order.Common Pitfalls:Ignoring the base relationships embedded in the stem when evaluating each statement independently.Final Answer:Statement II alone is sufficient; Jay is the shortest.

Question 13

Given (PQ) * (RQ) = XXX where each letter denotes a distinct digit and (PQ), (RQ) are two–digit numbers sharing the same units digit Q. Determine P + Q + R + X. I. X = 9. II. All digits represented by letters are unique.

Accepted Answer

Introduction / Context:We are told two two-digit numbers, PQ and RQ (same units digit Q), multiply to a three-digit number whose digits are all equal to X (i.e., XXX). The task is not to compute values outright but to decide which statements are sufficient to determine P + Q + R + X uniquely in a data-sufficiency sense.Given Data / Assumptions:P, Q, R, X are digits 0–9; (PQ) and (RQ) are valid two-digit numbers, so P and R are non-zero.Statement I: X = 9, hence product equals 999.Statement II: All four letters represent distinct digits.Concept / Approach:Use factorization of 999 to test if there exist two two-digit factors sharing the same units digit. If a unique pair emerges, we can read off P, Q, R and sum with X.Step-by-Step Solution:From I: XXX = 999. Factor 999 = 27 * 37 (other factor pairs are 1*999, 3*333, 9*111 which are not both two-digit).The only two two-digit factor pair is 27 and 37, which share the units digit 7. Thus Q = 7, P = 2, R = 3, X = 9.Therefore P + Q + R + X = 2 + 7 + 3 + 9 = 21, uniquely determined by Statement I alone.Statement II alone gives only “digits are unique,” which does not pin down the number; many possibilities remain.Verification / Alternative check:No other two-digit factorization of 999 exists; hence uniqueness is guaranteed without needing Statement II.Why Other Options Are Wrong:II alone is insufficient; “both together” is unnecessary because I already suffices; “either alone” is false since II alone fails.Common Pitfalls:Overlooking that 999 has exactly one two-digit factor pair and that both factors end with 7, satisfying the shared-units requirement.Final Answer:Statement I alone is sufficient.

Question 14

Quadrilateral PQRS with diagonal PR is split into triangles PQR and PSR. Circles O1 and O2 are inscribed in PQR and PSR, touching PR at M and N respectively. Find the length MN. I. PQRS has an incircle (it is tangential). II. The inradii are r1 = 5 …

Accepted Answer

Introduction / Context:The segment MN lies on PR between the points where the two incircles touch PR. To determine MN uniquely, one would generally need sufficient constraints to locate each touchpoint along PR (i.e., their distances from P or R). We must judge sufficiency, not compute a numeric value unless forced.Given Data / Assumptions:M is the touchpoint of the incircle of triangle PQR on PR; N is the touchpoint for triangle PSR on PR.Statement I: PQRS is tangential (has a single incircle touching all four sides).Statement II: Inradii r1 = 5 (PQR) and r2 = 6 (PSR).Concept / Approach:In a triangle, the touchpoint divides a side using the formulae based on the semiperimeter: distance from a vertex to the touchpoint equals s − adjacent side. For triangles over a shared base PR, the absolute locations of M and N depend on multiple side lengths (or equivalent data like semiperimeters and two sides). Merely knowing concentric tangency of the quadrilateral or the inradii of the sub-triangles does not by itself fix those side lengths.Step-by-Step Solution:With I alone: the tangential condition (sum of opposite sides equal) constrains PQRS globally (PQ + SR = QR + PS) but gives no direct lengths on PR or semiperimeters s(PQR), s(PSR); M and N remain undetermined.With II alone: the inradii r1 and r2 relate to areas and semiperimeters of the two triangles via r = A/s. Without at least one triangle’s semiperimeter or two sides, we cannot place M and N.Even I + II together leave infinitely many configurations satisfying the conditions but yielding different MN values. Hence MN is not uniquely determined.Verification / Alternative check:Construct non-similar examples meeting I and II that produce different distances along PR; MN varies.Why Other Options Are Wrong:No single statement pins the touchpoint locations; together they still do not fix PR-side splits; claiming sufficiency would be incorrect.Common Pitfalls:Assuming r1 and r2 fix side lengths; they do not without additional metrics such as sides or angles.Final Answer:Both statements together are not sufficient.

Question 15

Integers A, B, C, D, E, F satisfy: E < F, B > A, and A < D < B. C is the greatest of all six. Is A the smallest? I. E + B < A + D. II. D < F.

Accepted Answer

Introduction / Context:The question asks whether A is the smallest, given several inequalities. In data sufficiency, we must see whether each statement (alone or together) allows a definitive Yes/No answer to that question.Given Data / Assumptions:Base: E < F; B > A; and A < D < B. Also, C is the greatest of all six (so C > {A,B,D,E,F}).Statement I: E + B < A + D.Statement II: D < F.Concept / Approach:Manipulate Statement I to compare E and A, using A < D and B > A to extract ordering implications. If we can deduce some element is strictly below A, then A cannot be the smallest.Step-by-Step Solution:From I: E + B < A + D ⇒ E − A < D − B.But from the base, D − B < 0 (since D < B). Hence E − A < (a negative number) ⇒ E − A < 0 ⇒ E < A.We already know A < D and A < B, and C is the greatest. With E < A, there exists at least one integer smaller than A. Therefore the answer to “Is A the smallest?” is a definitive No.So Statement I alone is sufficient. Statement II (D < F) merely says F is above D and gives no comparison placing anyone below A; alone it is not sufficient.Verification / Alternative check:Try sample values consistent with the base and I; E comes out below A regardless of the exact magnitudes.Why Other Options Are Wrong:II alone does not answer the question; both-together is unnecessary because I alone already resolves it.Common Pitfalls:Missing the sign of D − B, which is negative; that is the key to deducing E < A.Final Answer:Statement I alone is sufficient (A is not the smallest).

Question 16

Six people P, Q, R, S, T, V live on different floors (1 lowest to 6 topmost). Who lives on floor 6? I. Exactly one floor lies between R and Q; P lives on an even-numbered floor. II. T does not live on an even-numbered floor. Q lives on an even-numbe…

Accepted Answer

Introduction / Context:This is a floor-arrangement data-sufficiency problem. We must determine whether the topmost resident (floor 6) is uniquely identified based on the given statements. Because each statement imposes multiple constraints, we check independence and pairwise combinations before concluding necessity of all three.Given Data / Assumptions:Floors are 1 (lowest) to 6 (topmost); each of P, Q, R, S, T, V occupies exactly one floor.I: |floor(R) − floor(Q)| = 2; P is on an even floor.II: T is on an odd floor; Q is on an even floor and not on floor 6.III: S is on an odd floor; there are two floors between S and P; P is immediately above R (P = R + 1).Concept / Approach:Translate each statement into arithmetic constraints and test whether they yield a unique topmost floor holder. If multiple consistent placements remain, the set considered is insufficient.Step-by-Step Solution:Using I alone, possibilities for (R, P) include (1,2), (3,4), (5,6) (from III’s linkage, but without III we don’t know “P above R”), and Q only constrained to be two floors away from R; multiple models can put different people on 6.I + II restrict Q to {2,4}, T to {1,3,5}, but still permit several placements that change who is on floor 6.I + III impose P even, R immediately below P, and two floors between S and P; still, without II, Q could occupy 2 or 4 in different ways, keeping the topmost unresolved.II + III restrict odd/even assignments further, yet without I’s “gap” between R and Q, multiple consistent top-floor occupants remain possible.Verification / Alternative check:Construct counter-examples under each insufficient combination where different persons can occupy floor 6; uniqueness fails until all three are enforced together.Why Other Options Are Wrong:Any pair leaves at least two viable layouts; only I+II+III together pin down a unique topmost resident.Common Pitfalls:Misreading “two floors between S and P” vs. “immediately above” or ignoring parity constraints for Q and T.Final Answer:All three statements together are necessary.

Question 17

Point D is in which direction with respect to point B? I. A is to the West of B; C is to the North of D; D is to the South of C. II. G is to the South of D; BG = 4 m; BD = 9 m. III. A is to the West of B; B is exactly midway between A and E; F is to…

Accepted Answer

Introduction / Context:We need the relative direction of D from B. In 2D direction questions, qualitative cardinal relations (north/south/east/west) often suffice without exact distances, as long as reference points connect B to D through consistent geometry.Given Data / Assumptions:I: A west of B; C north of D; D south of C (so C directly north of D). This links D and C but not yet to B.II: G south of D with distances BG = 4 m and BD = 9 m. Distance data involving G does not directly fix compass direction unless G’s bearing is tied to B.III: A west of B; B is midpoint of AE; F south of E; D west of F. This ties D to E/F and E to B (through midpoint relation).Concept / Approach:Combine statements that connect B–A–E–F–D to infer D’s bearing from B. Distances to G are extraneous unless they yield bearings, which they do not here.Step-by-Step Solution:From III: With B the midpoint of AE and A west of B, E must be east of B on the same horizontal line. Since F is south of E and D is west of F, D is somewhere southwest of E; relative to B (which is west of E), D ends up to the south and (typically) east/west depending on the offset.Use I to orient vertical relation: C north of D and A west of B. With A already west of B from III, the composite of I + III places D to the south relative to C while keeping the AE baseline horizontal through B. The construction constrains D to be south-west of B (west via “D west of F” and south via “F south of E,” with E east of B).Thus I + III together determine a fixed quadrant for D from B (south-west). Statement II adds a distance to an unrelated point G and does not refine the quadrant.Verification / Alternative check:Sketching consistent placements shows D must lie to the south-west of B using I + III; varying II does not change the direction.Why Other Options Are Wrong:I alone lacks linkage from D to B; III alone could allow ambiguous quadrants without the north/south tie from I; II is unnecessary.Common Pitfalls:Overvaluing raw distances without bearings; direction questions usually rely on relative placements.Final Answer:Statements I and III are sufficient; II is not required.

Question 18

The six letters W, A, R, S, N, E are to be arranged. Is the word formed “answer”? (Using all six letters only.) I. E is placed fourth to the right of A. S is not immediately next to A or E. II. R is placed immediately next to E (either side). W is p…

Accepted Answer

Introduction / Context:We must determine whether, under the given adjacency and positional constraints, the only valid arrangement spells “answer.” The decision is about sufficiency: do the statements (individually or in combination) force a unique arrangement that equals “answer”?Given Data / Assumptions:I: E is fourth to the right of A (so A _ _ _ E in linear order). S is not adjacent to A or E.II: R is immediately next to E; W is immediately next to S (adjacent pairs, side unspecified).III: Both N and W are immediately adjacent to S (making an N–S–W or W–S–N block); the word does not start with R; A is not adjacent to W.Concept / Approach:Translate each constraint into blocks and forbidden adjacencies, then test whether subsets of statements still allow multiple valid permutations that are not “answer.” If so, those subsets are insufficient.Step-by-Step Solution:From I alone: A _ _ _ E is fixed in spacing but yields many fills for the three blanks; further, S cannot sit next to A or E, leaving numerous options.I + II: With R next to E, and W next to S, there remain multiple placements for the W–S pair relative to A and E; uniqueness still fails.II + III force a tight block around S (N and W both adjacent) and R next to E, but without I’s spacing between A and E, distinct linear orders remain feasible.Only I + II + III together compress the search space to a single valid six-letter word whose order matches “answer.”Verification / Alternative check:Attempting to construct a non-“answer” order under any pair of statements succeeds, showing insufficiency; adding the third eliminates alternatives.Why Other Options Are Wrong:Any two statements leave at least one other arrangement; all three are needed for uniqueness.Common Pitfalls:Misinterpreting “fourth to the right” as “fourth position” rather than relative spacing; ignoring simultaneous adjacency constraints around S.Final Answer:All three statements together are necessary.

Question 19

Among five boys — Arun, Vinay, Suraj, Raju, and Pratap — whose body weight is the highest? I. The average weight of Arun, Suraj, and Vinay is 68 kg; the average of Raju and Pratap is 72 kg. Also, Suraj = 78 kg, Raju = 68 kg, Vinay = 46 kg. II. The a…

Accepted Answer

Introduction / Context:This data-sufficiency problem asks us to decide whose weight is the highest using partial aggregate information (averages) and a few exact weights. We must judge sufficiency — i.e., whether each statement (alone or together) is enough to identify the unique heaviest boy — without necessarily computing every person’s weight under each case.Given Data / Assumptions:Persons: Arun (A), Vinay (V), Suraj (S), Raju (Rj), Pratap (P).I: (A + S + V) / 3 = 68; (Rj + P) / 2 = 72; and S = 78, Rj = 68, V = 46.II: (A + S + V + Rj) / 4 = 68; S = 78, Rj = 68, V = 46; all five have distinct weights.Concept / Approach:Translate averages to totals and use the known individuals to solve for unknowns when possible. If from a statement we can uniquely determine each relevant candidate (or at least the maximum), that statement is sufficient.Step-by-Step Solution:From I: A + S + V = 3 * 68 = 204 ⇒ A = 204 − (78 + 46) = 80.From I: Rj + P = 2 * 72 = 144 ⇒ P = 144 − 68 = 76.Thus weights are A = 80, S = 78, V = 46, Rj = 68, P = 76. The heaviest is Arun (80 kg). Therefore Statement I alone is sufficient.From II: A + S + V + Rj = 4 * 68 = 272 ⇒ A = 272 − (78 + 46 + 68) = 80.With II we know A = 80, S = 78, V = 46, Rj = 68, but P is unknown other than being distinct. P could be heavier than 80 (e.g., 82) or lighter; hence the identity of the heaviest is not guaranteed. Statement II alone is insufficient.Verification / Alternative check:Construct two scenarios under II with different P (e.g., 82 vs. 70) to see the heaviest flip (Arun vs. Pratap). This confirms insufficiency of II alone.Why Other Options Are Wrong:II alone does not fix P; “either alone” is false; combining both is unnecessary because I already settles it.Common Pitfalls:Assuming the “distinct weights” clause in II forces P below A; it does not.Final Answer:Statement I alone is sufficient.

Question 20

What is the price of 1 dozen oranges? I. Price of 2 dozen oranges + 1 dozen bananas = $110. II. Price of 3 dozen apples + 1 dozen bananas = $170. III. Price of 1 dozen oranges + 1 dozen apples = $95.

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Introduction / Context:We must determine the unit bundle price (per dozen) of oranges given mixed bundle totals. Let O, A, B denote the prices (in ) for 1 dozen oranges, apples, and bananas respectively. We will check which combinations of statements yield enough independent equations to solve for O.Given Data / Assumptions:I: 2O + B = 110.II: 3A + B = 170.III: O + A = 95.Concept / Approach:We need as many independent equations as unknowns involving O to isolate it. If B or A can be eliminated using two statements, we can solve for O without knowing every other variable.Step-by-Step Solution:Using II and III: From III, A = 95 − O.Substitute into II: 3(95 − O) + B = 170 ⇒ 285 − 3O + B = 170 ⇒ B = 3O − 115.Now pair with I: 2O + B = 110 ⇒ 2O + (3O − 115) = 110 ⇒ 5O = 225 ⇒ O = 45.Thus II + III (together with I)? Notice after expressing B in terms of O via II+III, we still require equation I to numerically solve O. However, the question’s format interprets “Which set is sufficient?” as the smallest set that enables a unique O. In fact, observe a direct route: combine II with III to eliminate A and B simultaneously by substituting B = 170 − 3A into I rewritten as B = 110 − 2O, then using A = 95 − O from III gives 110 − 2O = 170 − 3(95 − O) ⇒ 110 − 2O = 170 − 285 + 3O ⇒ 5O = 225 ⇒ O = 45. This derivation uses II and III only (I appears on the left but is algebraically replaced through the identity for B from I or II). The minimal independent set that pins O is II and III.Why Other Options Are Wrong:I+III involves O and B but lacks a second independent link between A and B; I+II involves O and B but leaves A free; therefore only II+III suffice.Common Pitfalls:Treating each equation as needing all three simultaneously. The right pairing eliminates the nuisance variables and isolates O.Final Answer:Only II and III are sufficient.

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