CSAT Solved Papers/ 2022/Q15

2022 CSAT — Q15

Quant Logical & quantitative reasoning 2.5 marks Medium

$X$ and $Y$ run a $3$ km race along a circular course of length $300$ m. Their speeds are in the ratio $3 : 2$ . If they start together in the same direction, how many times would the first one pass the other (the start-off is not counted as passing)?

A 2
B 3 Answer
C 4
D 5

Worked rationale

The race ends when the faster runner $X$ completes $3$ km. In that time, with speeds $3:2$ , $Y$ covers $\tfrac{2}{3}\times 3 = 2$ km (and is still on the course).

$X$ passes (laps) $Y$ each time $X$ gains one full lap, $300$ m, on $Y$ . The relative gain of $X$ over $Y$ by the finish is

3000 - 2000 = 1000 \text{ m}.

Number of laps gained $= \left\lfloor \dfrac{1000}{300} \right\rfloor = \lfloor 3.33 \rfloor = 3$ . The passes occur at relative gains of $300, 600, 900$ m — all reached while both runners are still running ( $Y$ finishes $2$ km only after $X$ finishes). So $X$ passes $Y$ three times.

Answer: (b) 3.

Why the other options miss

A

counted one too few: stops at a relative gain of $600$ m, missing the third pass at $900$ m (still $< 1000$ ).
C

counted one too many: counts the $1000$ -m mark as a fourth pass, but $1000/300 = 3.33$ — no fourth full lap is gained.
D

wrong formula: counts total laps ( $X$ ‘s $10$ ) or sums laps instead of relative laps.

Specialist insight

The passing count is governed by relative distance, not total distance: $X$ overtakes $Y$ once per $300$ m of relative gain, and the gain at the finish is $1000$ m $\Rightarrow \lfloor 1000/300\rfloor = 3$ . The “start-off not counted” clause is the warning that you take the floor of the relative-laps, not $\lceil\cdot\rceil$ , and you do not credit the starting coincidence.

The trap, in one line

Relative gain at finish $= 1000$ m; passes $= \lfloor 1000/300 \rfloor = 3 \Rightarrow$ (b).

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