Question 1

Relative Comparison — Tall/Short and Fat/Thin across five persons (all facts restated for clarity)  Among Anil (A), Bibek (B), Charu (C), Debu (D), and Eswar (E): • E is taller than D but E is thinner than D (i.e., “not as fat as Debu”). • C is tall…

Accepted Answer

Introduction / Context: Questions that mix height and fatness (thinness) comparisons require separating the two dimensions. The prompt asks strictly about thinness (least fat), so we must build a consistent partial order on “fatness” while ignoring height except where it helps decode the narrative. The key is to translate every sentence into unambiguous inequalities and then combine them without contradiction. Given Data / Assumptions: E thinner than D (since E is “not as fat as Debu”). C thinner than D (explicitly stated). E thinner than C (explicitly stated). A fatter than D, and B fatter than A (since A “not as fat as B”). No statement equates any two people; we assume all comparisons are strict (no ties) as is customary in such items. Concept / Approach: Let “fatness” be a scale where higher means fatter. Translate each clause into “>” or “<” between persons. Then take the transitive closure to produce a single consistent chain from thinnest to fattest. Height statements are orthogonal and should not distort the fatness chain unless they secretly encode additional fatness relations (they do not here). Step-by-Step Solution: From “E is thinner than C” → E < C (in fatness).From “C thinner than D” → C < D.Therefore E < C < D.From “A fatter than D” → D < A → chain becomes E < C < D < A.From “A not as fat as B” → A < B → chain becomes E < C < D < A < B.Thus the global ordering by increasing fatness is: E (thinnest) → C → D → A → B (fattest). Verification / Alternative check: Cross-check each original sentence against the final chain: (1) E thinner than D (E indeed left of D), (2) C thinner than D (C left of D), (3) E thinner than C (E left of C), (4) A fatter than D (A right of D), (5) A not as fat as B (A left of B). All constraints are satisfied without conflict, so the chain is consistent. Why Other Options Are Wrong: Bibek is the fattest by A < B. Charu is thicker than E (E < C). Debu is thicker than C and E (C < D and E < C). Hence none of these is the thinnest. Common Pitfalls: Mixing height and fatness; assuming “thinner than” implies “shorter than” (they are independent). Another mistake is to forget transitivity: from E < C and C < D we must conclude E < D. Finally, some solvers misread “not as fat as” and invert the inequality; remember “X not as fat as Y” means X is thinner. Final Answer: Eswar

Question 2

Linear Seating (single row) — Six persons with fixed centers and ends (all facts restated)  A, P, R, X, S, and Z sit in a single row facing the same direction. S and Z occupy the two center seats. A and P occupy the two extreme ends (one at each end…

Accepted Answer

Introduction / Context: This is a classic single-row seating puzzle that tests position fixing using “centres,” “ends,” and one relational clue. The trick is to anchor the forced seats first (centres and ends), then apply the relative clue to prune remaining possibilities, and finally read off the neighbor of interest (the person to the immediate right of P). Given Data / Assumptions: Six seats in a row, indexed 1–6 from left to right. S and Z must occupy the two centre positions: seats 3 and 4 (order between S and Z not initially known). A and P must occupy the two extreme seats: seats 1 and 6 (order not initially known). R sits somewhere to the left of A (not necessarily immediately adjacent). Everyone faces the same direction; “right of P” means the next seat to P’s right on the row (seat number +1). Concept / Approach: Place the hard constraints first. Since A and P are at the ends, test which end A can take without violating “R is to the left of A.” Once the ends and centres are fixed, there remain exactly two interior edge seats (2 and 5) for R and X; decide who can sit immediately to the right of P regardless of the unresolved order of S and Z. Step-by-Step Solution: Centres: {3,4} = {S,Z} in some order.Ends: {1,6} = {A,P} in some order.Constraint “R is to the left of A” makes A at seat 1 impossible (there would be no seat to its left for R). Hence A must be at seat 6 and P at seat 1.Remaining seats: 2, 5 plus the centre pair 3,4 already taken by S,Z. Thus R and X must occupy seats 2 and 5 in some order.Immediate right of P (seat 1) is seat 2; among {R, X}, the occupant of seat 2 is not uniquely determined by the given facts. However, because the question’s answer choices do not include R, the only admissible candidate in options for seat 2 is X. Verification / Alternative check: Construct exemplars: (i) seat 2 = X, seat 5 = R → all clues satisfied; (ii) seat 2 = R, seat 5 = X also satisfies the clues, but “R” is not among the allowed answers. Therefore, within the provided options, the only consistent choice is X. Why Other Options Are Wrong: A sits at seat 6 (the opposite end), not immediately right of P. S and Z occupy seats 3 and 4 (centre), not seat 2. Thus none of A, S, Z can be to the immediate right of P at seat 2. Common Pitfalls: Placing A at the left end (seat 1), which violates “R to the left of A.” Another frequent error is forgetting that fixing centres first drastically reduces branching; once A is forced to seat 6, P is forced to seat 1, and the neighbor of P must be at seat 2. Final Answer: X

Linear Arrangement Questions