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(D) $1$. Let the upward journey sequence be $1, 2, 3, 4, 5$.$2$. $D$ is the first stop in the downward journey,so $D$ is the last stop $(5)$ in the up

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(D) $1$. Let the upward journey sequence be $1, 2, 3, 4, 5$.$2$. $D$ is the first stop in the downward journey,so $D$ is the last stop $(5)$ in the upward journey.$3$. $A$ and $E$ are not terminal stops,so they cannot be $1$ or $5$. Thus,$A, E \in \{2, 3, 4\}$.$4$. $C$ is not the middle stop $(3)$,so $C \in \{1, 2, 4\}$.$5$. The condition '$C$ comes twice as many stops before $D$ as $B$ comes after $A$' implies: Let $x$ be the number of stops between $A$ and $B$ (where $B$ is after $A$),and $y$ be the number of stops between $C$ and $D$ (where $C$ is before $D$). Then $y = 2x$.$6$. If $A=2$ and $B=3$,then $x=0$. This would mean $y=0$,so $C$ is immediately before $D$. If $D=5$,then $C=4$. Remaining stops $E=1$ (not possible,$E$ is not terminal).$7$. If $A=2$ and $B=4$,then $x=1$. Then $y=2$. If $D=5$,then $C$ is $2$ stops before $D$,so $C=5-2-1 = 2$ (not possible,$A=2$).$8$. If $A=3$ and $B=4$,then $x=0$. Then $y=0$. $C$ is immediately before $D=5$,so $C=4$. Remaining stops $E=2$. Sequence: $1-B, 2-E, 3-A, 4-C, 5-D$. Wait,$B$ is after $A$,so $A=3, B=4$ works. $C=4$ is not possible as $B=4$. $9$. Let's re-evaluate: Upward sequence $1-B, 2-E, 3-A, 4-C, 5-D$. $A$ is at $3$,$B$ is at $1$ (not after $A$).$10$. Correct sequence: $1-B, 2-A, 3-E, 4-C, 5-D$. $A=2, B=1$ (No). $11$. Correct sequence: $1-E, 2-A, 3-B, 4-C, 5-D$. $A=2, B=3$ $(x=0)$,$C=4, D=5$ $(y=0)$. $C$ is not middle $(3)$. $A, E$ not terminals. This works. $12$. Downward journey is the reverse of upward: $D, C, B, A, E$.

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