CSAT Solved Papers/ 2025/Q65

2025 CSAT — Q65

Quant Logical & quantitative reasoning 2.5 marks Hard

There are $7$ places $A, B, C, D, E, F$ and $G$ in a city connected by various roads $AB, AC, CD, DE, BF, EG$ and $FG$ . $A$ is $6$ km south of $B$ . $A$ is $10$ km west of $C$ . $D$ is $5$ km east of $E$ . $C$ is $6$ km north of $D$ . $F$ is $9$ km west of $B$ . $F$ is $12$ km north of $G$ . A person travels from $D$ to $F$ through these roads. What is the distance covered by the person?

A 20 km
B 25 km
C 31 km Answer
D 37 km

Worked rationale

The seven places are joined only by the listed roads: $AB, AC, CD, DE, BF, EG, FG$ . A trip from $D$ to $F$ must follow these edges, and the marks come from summing the known road lengths along a valid route.

Roads with given lengths: $AB = 6,\ AC = 10,\ CD = 6,\ DE = 5,\ BF = 9,\ FG = 12$ (the length of $EG$ is not given). Trace a path from $D$ to $F$ using only measured roads:

D \xrightarrow{CD = 6} C \xrightarrow{AC = 10} A \xrightarrow{AB = 6} B \xrightarrow{BF = 9} F.

\text{Distance} = 6 + 10 + 6 + 9 = 31\ \text{km}.

(The alternative $D \to E \to G \to F$ uses $EG$ , whose length is unspecified, so it is not the intended computable route.)

Answer: (c) 31 km.

Visual solution

The same solve, worked by hand — read it, then trace it.

Hand-drawn worked solution for UPSC 2025 CSAT Q65 — Logical & quantitative reasoning — Tap the drawing to open it full size for the fine detail.

Why the other options miss

A

missed a leg of the route: sums only part of the route (e.g. $CD+AC+\dots$ dropping a leg), short by the $AB$ or $BF$ segment.
B

an arithmetic slip: in adding $6+10+6+9$ (e.g. uses $AC=4$ or drops a unit).
D

solved the wrong question: adds an extra road not on the $D$ – $F$ path (double-counts a leg or includes $DE$ ).

Specialist insight

This is a graph-traversal item dressed as a directions puzzle: the directions only confirm the road lengths; the real task is to route $D \to F$ along the given edges and add the lengths. The trap is the $D\!-\!E\!-\!G\!-\!F$ alternative — it looks shorter on a sketch, but $EG$ has no stated length, so it cannot be the keyed route. Identify which roads carry numbers first, then find the $D$ -to- $F$ path that uses only those: $D\!-\!C\!-\!A\!-\!B\!-\!F = 31$ . A quick coordinate sketch (place $A$ at origin) confirms the adjacencies without ambiguity.

The trap, in one line

Route $D\!\to\!C\!\to\!A\!\to\!B\!\to\!F$ along the measured roads: $6+10+6+9 = 31$ km (the $E$ – $G$ route is unusable — $EG$ 's length isn't given).

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