UPSC CDS 2 2023 Math Paper

Concept:The distance is constant; the change in speed causes a change in time, allowing us to set up an equation for the original travel time.Explanation:Let the original time to reach the office be $T$ minutes.When walking at $3$ km/hr (speed in km per hour), the man reaches $40$ minutes early, so the actual time taken is $(T - 40)$ minutes.When walking at $2$ km/hr, he reaches $40$ minutes late, so the actual time taken is $(T + 40)$ minutes.Since $\text{distance} = \text{speed} \times \text{time}$, and the distance is the same in both cases, we convert minutes to hours by dividing by $60$.Thus, $3 \times \frac{(T - 40)}{60} = 2 \times \frac{(T + 40)}{60}$ Multiplying both sides by $60$:$3 (T - 40) = 2 (T + 40)$Expanding: $3T - 120 = 2T + 80$Solving: $3T - 2T = 80 + 120$ → $T = 200$ minutes.Now, distance $D = 3 \times \frac{(200 - 40)}{60} = 3 \times \frac{160}{60} = 3 \times \frac{8}{3} = 8$ km.Answer:$8$ km (Option B).