2025 CSAT — Q39

To completely pass a walker, the tram covers its own length $L$ at the relative speed (same direction $\Rightarrow$ subtract). Let the tram’s speed be $v$ km/hr. Then in m/s the relative speeds are $(v-3)\tfrac{5}{18}$ and $(v-4)\tfrac{5}{18}$, and

Cancel the common $\tfrac{5}{18}$ and solve:

Substitute back:

(Cross-check with $Y$: $(12-4)\cdot\tfrac{5}{18}\cdot 9 = 8\cdot\tfrac{5}{18}\cdot 9 = 20$ ✓.)

Answer: (c) 20 m.

The single fact that organises every “train passes a moving object” item: distance covered = own length, speed used = relative speed. Same direction subtracts, opposite adds — getting this sign wrong is distractor (d). The elegant move here is to cancel the unit-conversion factor $\tfrac5{18}$ before solving — it appears on both sides, so $8(v-3) = 9(v-4)$ needs no m/s arithmetic at all, and $v$ falls out cleanly. Only at the final substitution do you convert once. That ordering (algebra first, conversion last) is both faster and slip-proof under the clock.