2022 CSAT — Q30

Number positions from the front. ”$11$ persons ahead of $C$” fixes $C$ at position $12$. ”$13$ behind $A$” gives total $N = (\text{position of } A) + 13$, so minimising $N$ means putting $A$ as far forward as possible.

Gaps: ”$4$ between $A$ and $B$” $\Rightarrow |A - B| = 5$; ”$7$ between $B$ and $C$” $\Rightarrow |B - C| = 8$. With $C = 12$, $B = 12 \pm 8 = 4$ or $20$.

To push $A$ forward, take $B = 4$ (the front option). Then $A = B \pm 5 = 9$ or $-1$; the valid front-most is $A = 9$.

Check $A=9, B=4, C=12$: between $A$ and $B$ are positions $5,6,7,8$ — exactly $4$ persons ✓; between $B$ and $C$ are $5,6,7,8,9,10,11$ — exactly $7$ persons ✓ (the people in the gaps may overlap, which is allowed). Then

The alternative $B=20$ forces $A \ge 15$, giving $N \ge 28$ — larger.

Answer: (a) 22.

Minimum-queue items are won by letting the gap-regions overlap. Fix the anchored person ($C$ at $12$), choose the gap directions that crowd everyone toward the front, and verify each “between” count literally. Here $A=9, B=4, C=12$ packs all constraints with the between-$B$-and-$C$ block swallowing $A$, yielding $N=22$ — far below the no-overlap stack of $38$.