2022 CSAT — Q6

Quant Logical & quantitative reasoning 2.5 marks Medium

Two friends $X$ and $Y$ start running and they run together for $50$ m in the same direction and reach a point. $X$ turns right and runs $60$ m, while $Y$ turns left and runs $40$ m. Then $X$ turns left and runs $50$ m and stops, while $Y$ turns right and runs $50$ m and then stops. How far are the two friends from each other now?

A 100 m Answer
B 90 m
C 60 m
D 50 m

Worked rationale

Put the common start at the origin and the shared first leg pointing North. Track coordinates $(x,y)$ .

Both run North $50$ : reach $(0,50)$ .

$X$ : turns right (faces East) runs $60 \Rightarrow (60,50)$ ; turns left (faces North) runs $50 \Rightarrow (60,100)$ .

$Y$ : turns left (faces West) runs $40 \Rightarrow (-40,50)$ ; turns right (faces North) runs $50 \Rightarrow (-40,100)$ .

Both end at height $y = 100$ , so the separation is purely horizontal:

|60 - (-40)| = 100 \text{ m}.

Answer: (a) 100 m.

Visual solution

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

Hand-drawn worked solution for UPSC 2022 CSAT Q6 — Logical & quantitative reasoning — Tap the drawing to open it full size for the fine detail.

Why the other options miss

B

an arithmetic slip: mis-adds the horizontal legs (e.g. $60 + 40$ but then nets a $10$ from a phantom vertical mismatch).
C

solved the wrong question: uses only $X$ ‘s eastward leg, forgetting $Y$ moved $40$ m the other way.
D

solved the wrong question: answers a single leg length instead of the final separation.

Specialist insight

The killer simplification is noticing both walkers finish at the same northward height ( $+100$ ): their final vertical legs ( $X$ : North $50$ ; $Y$ : North $50$ ) leave them level, so the distance is just the sum of the opposite horizontal excursions $60 + 40 = 100$ . Coordinate-tracking beats mental rotation every time on turn-by-turn direction items — write the points, don’t picture them.

The trap, in one line

Both end at $y=100$ ; horizontal split $= 60 + 40 = 100$ m $\Rightarrow$ (a).

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