SBI PO Prelims 8 Mar 2025 Shift 1 Paper

Concept:This is a positional puzzle with nine boxes ($O$, $P$, $Q$, $R$, $S$, $T$, $U$, $V$, $W$) placed from position $1$ (bottom) to $9$ (top). We need to find which box is at the $6^{\text{th}}$ position.Explanation:• Condition 1: Four boxes between $T$ and $U$. More than four boxes above $U$, and $U$ is even-numbered.So $|\text{pos}(T) - \text{pos}(U)| = 5$ and $\text{pos}(U) < 5$. Hence $U$ can be $2$ or $4$.If $U=2$, then $T=7$. If $U=4$, then $T=9$.• Condition 2: Two boxes between $O$ and $P$. $P$ is just above $U$ (so $\text{pos}(P) = \text{pos}(U) + 1$).If $U=2$: $P=3$, then $|\text{pos}(O)-3|=3$ gives $O=6$.If $U=4$: $P=5$, then $O$ could be $2$ or $8$.• Condition 3: Three boxes between $S$ and $R$. $R$ is just above $T$.For $U=2$: $T=7$ so $R=8$. Then $|\text{pos}(S)-8|=4$ gives $S=4$.For $U=4$: $T=9$ so $R=10$ (impossible). So only $U=2$ works.So far: $U=2$, $P=3$, $S=4$, $O=6$, $T=7$, $R=8$. Free positions: $1,5,9$. Remaining boxes: $V,W,Q$.• Condition 4: More than three boxes between $P$ and $V$. $V$ is above $O$ ($\text{pos}(V) > 6$).Only free position $>6$ is $9$. Check boxes between $P(3)$ and $V(9)$: five boxes, so $V=9$.• Condition 5: More than two boxes between $W$ and $Q$. $Q$ is not above $P$, so $\text{pos}(Q) \le 3$.Free $\le3$ only $1$ (since $2,3$ taken). So $Q=1$. Then $|\text{pos}(W)-1|-1 > 2$ gives $\text{pos}(W) \ge 5$. Free $5$ works: boxes between $W(5)$ and $Q(1)$ are $2,3,4$ (three boxes) so $W=5$.Final arrangement (position → box): $1→Q$, $2→U$, $3→P$, $4→S$, $5→W$, $6→O$, $7→T$, $8→R$, $9→V$.The $6^{\text{th}}$ position holds box $O$.Answer:O