Question:

Six students, Arif (Ar), Balwinder (Bw), Chintu (Ct), David (Dv), Emon (Em) and Fulmoni (Fu) appeared in GATE–XH (2022). Bw scores less than Ct in XH–B1, but more than Ar in XH–C1. Dv scores more than Bw in XH–C1, and more than Ct in XH–B1. Em scores less than Dv, but more than Fu in XH–B1. Fu scores more than Dv in XH–C1. Ar scores less than Em, but more than Fu in XH–B1. Who scores highest in XH–B1?

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In order-comparison puzzles, segregate constraints by category (here, by paper B1 vs C1), write only the relevant inequalities, then build a single chain. A strict "$>$" link from the topper to each other candidate certifies the topper.
Updated On: Aug 29, 2025
  • Fulmoni
  • Emon
  • David
  • Chintu
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The Correct Option is C

Solution and Explanation


 

Step 1: Collect XH–B1 inequalities.
From the statements for paper B1: \[ \begin{aligned} \text{(i)}\ &\text{Bw} < \text{Ct} \text{(ii)}\ &\text{Dv} > \text{Ct} \text{(iii)}\ &\text{Em} < \text{Dv}, \text{Em} > \text{Fu} \text{(iv)}\ &\text{Ar} < \text{Em}, \text{Ar} > \text{Fu} \end{aligned} \]

Step 2: Chain what we can.
From (iv) and (iii): $\text{Fu} < \text{Ar} < \text{Em} < \text{Dv}$. From (ii): $\text{Ct} < \text{Dv}$. From (i) and (ii): $\text{Bw} < \text{Ct} < \text{Dv}$.

Step 3: Decide the topper in B1.
Every candidate is strictly below Dv: - $\text{Ct} < \text{Dv}$ (given), hence $\text{Bw} < \text{Ct} < \text{Dv}$.
- $\text{Em} < \text{Dv}$ (given), and $\text{Ar} < \text{Em}$, $\text{Fu} < \text{Ar}$. 

Therefore, \(\boxed{\text{Dv is the highest in XH–B1}}\).

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