RRB Group D 1 Dec 2025 Shift 2 Solved Paper

Concept:Circular arrangement where all face centre: left = anticlockwise, right = clockwise.Explanation:Step 1: Place $C$ at an arbitrary seat $1$.$B$ sits immediate right (clockwise) of $C$ → $B$ at seat $2$.Step 2: From left (anticlockwise) of $B$, exactly three persons lie between $B$ and $A$.Moving anticlockwise from $B$: first person is $C$ (seat $1$), then seats $7$, $6$, then $A$ at seat $5$.Thus $A$ is $4$ steps anticlockwise from $B$ → $A$ at seat $5$.Step 3: From left of $C$, exactly three persons lie between $C$ and $D$.Anticlockwise from $C$ (seat $1$): first person seat $7$, then seat $6$, then seat $5$ ($A$), then $D$ at seat $4$.Thus $D$ at seat $4$.Step 4: Occupied seats: $1$: $C$, $2$: $B$, $4$: $D$, $5$: $A$.Remaining seats: $3$, $6$, $7$ for $E$, $F$, $G$.Step 5: $G$ sits immediate right of $F$ → $F$ then $G$ clockwise.Place $F$ at seat $6$, $G$ at seat $7$ (clockwise from $6$). Then $E$ occupies seat $3$.Step 6: Final clockwise order: $1$:$C$, $2$:$B$, $3$:$E$, $4$:$D$, $5$:$A$, $6$:$F$, $7$:$G$.Verify: Left of $B$ to $A$ anticlockwise: $B$→$C$→$G$→$F$→$A$ → three persons ($C,G,F$).Left of $C$ to $D$ anticlockwise: $C$→$G$→$F$→$A$→$D$ → three persons ($G,F,A$).Immediate right of $F$: $G$ clockwise, correct.Step 7: Count people between $A$ and $C$ from right of $C$ (clockwise).From $C$ clockwise: $B$, $E$, $D$, then $A$.Persons between: $B$, $E$, $D$ → $3$ persons.Answer:$3$