CSAT Solved Papers/ 2025/Q37

2025 CSAT — Q37

Quant Logical & quantitative reasoning 2.5 marks Hard

$P$ and $Q$ walk along a circular track. They start at $5{:}00$ a.m. from the same point in opposite directions. $P$ walks at an average speed of $5$ rounds per hour and $Q$ walks at an average speed of $3$ rounds per hour. How many times will they cross each other between $5{:}20$ a.m. and $7{:}00$ a.m.?

A 12
B 13
C 14 Answer
D 15

Worked rationale

Walking in opposite directions, their meeting rate is the sum of speeds:

5 + 3 = 8 \text{ meetings per hour.}

Starting together at $5{:}00$ , a crossing occurs each time their combined rounds total an integer, i.e. every $\tfrac{1}{8}$ hour $= 7.5$ minutes. Crossings happen at $5{:}00 + 7.5k$ minutes for $k = 1, 2, 3, \dots$

Count the $k$ with crossing-time in the window $(5{:}20,\ 7{:}00]$ , i.e. $7.5k$ minutes in $(20, 120]$ :

7.5k > 20 \Rightarrow k \ge 3 \quad (k=3 \to 22.5\text{ min}),

7.5k \le 120 \Rightarrow k \le 16 \quad (k=16 \to 120\text{ min} = 7{:}00).

So $k = 3, 4, \dots, 16$ — that is $16 - 3 + 1 = 14$ crossings. (At $5{:}20$ exactly, $8 \times \tfrac{20}{60} = 2.\overline{6}$ is not an integer, so $5{:}20$ is not itself a crossing; at $7{:}00$ , $8 \times 2 = 16$ is, so the endpoint counts.)

Answer: (c) 14.

Visual solution

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

Hand-drawn worked solution for UPSC 2025 CSAT Q37 — Logical & quantitative reasoning — Tap the drawing to open it full size for the fine detail.

Why the other options miss

A

the proportion runs the wrong way: uses the difference $5-3=2$ (the same-direction overtaking rate) somewhere, or measures the full $5{:}00$ – $7{:}00$ minus a chunk, undershooting.
B

off by one: excludes the $7{:}00$ endpoint crossing (treats the window as open), giving $k=3..15$ .
D

off by one: includes a phantom crossing at or before $5{:}20$ (counts $k=2$ ), overshooting by one.

Specialist insight

Two things decide this item. First, opposite directions ⇒ add the speeds (meetings per hour $=5+3=8$ ); using the difference is the classic same-vs-opposite confusion. Second, it is a window-counting problem, not a duration: convert to “crossing instants every $7.5$ min from $5{:}00$ ” and count integers $k$ with $7.5k \in (20,120]$ . The endpoints are where marks are lost — $5{:}20$ is not a crossing, $7{:}00$ is — so anchoring both boundaries explicitly is the discipline that separates $14$ from the off-by-one decoys $13$ and $15$ .

The trap, in one line

Opposite directions give $5+3=8$ meetings/hour (every $7.5$ min from $5{:}00$ ); counting $k$ with $7.5k\in(20,120]$ gives $k=3..16 = 14$ .

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