Question

A bus starts at 9 am and follows a fixed route every day. One day, it traveled at a constant speed of 60 km per hour and reached its destination 3.5 hours later than its scheduled arrival time. Next day, it traveled two-thirds of its route in one-third of its total scheduled travel time, and the remaining part of the route at 40 km per hour to reach just on time. The scheduled arrival time of the bus is

Accepted Answer

C. Let total distance = $D$ km, and scheduled time = $T$ hours. From Day 1: $$D = 60(T + 3.5)$$ From Day 2: First $\frac{2}{3}D$ is traveled in time $\frac{1}{3}T$. Remaining distance $\frac{1}{3}D$ is traveled at $40	ext{ km/h}$ in remaining time $\frac{2}{3}T$. $$\frac{\frac{1}{3}D}{40} = \frac{2}{3}T \implies D = 80T$$ Equating both expressions for $D$: $$80T = 60(T + 3.5) \implies 20T = 210 \implies T = 10.5	ext{ hours}$$ Departure time = 9:00 am. Scheduled arrival time = $9:00	ext{ am} + 10.5	ext{ hours} = 7:30	ext{ pm}$.

Scheduled Arrival Time of Bus

Explanation

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