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
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
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
Options
Answer: 7 : 30 pm
Explanation:
Let the total distance = 3d Let the total scheduled time = 3t Therefore, speed = 3d / 3t = d / t Given: d / (2t) = 40 km/hr ⇒ d / t = 80 km/hr Total distance = 80 × 3t Also given the delay condition: 80 × 3t = 60 × (3t + 3.5) 240t = 180t + 210 60t = 210 t = 3.5 So, total time = 3t = 3 × 3.5 = 10.5 hours Scheduled arrival time = 9:00 AM + 10.5 hours = 7:30 PM
Explanation:
Let the total distance = 3d Let the total scheduled time = 3t Therefore, speed = 3d / 3t = d / t Given: d / (2t) = 40 km/hr ⇒ d / t = 80 km/hr Total distance = 80 × 3t Also given the delay condition: 80 × 3t = 60 × (3t + 3.5) 240t = 180t + 210 60t = 210 t = 3.5 So, total time = 3t = 3 × 3.5 = 10.5 hours Scheduled arrival time = 9:00 AM + 10.5 hours = 7:30 PM
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