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 taller than A but shorter than B (so, by height: B > C > A among these three). • A is fatter than D but not as fat as B (so, by fatness: B is fatter than A, and A is fatter than D). • E is thinner than C, and C is thinner than D (hence D is fatter than C, and C is fatter than E). Question: Who is the thinnest person (i.e., least fat) among A, B, C, D, E?

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