A bus starts at 8 am and follows a fixed route every day. One day, it travels at a constant speed of 60 km/hr and reaches its destination 3 hours later than its scheduled arrival time. The next day, it travels two-thirds of its route in one-third of its total scheduled travel time, and the remaining part of the route at 40 km/hr to reach exactly on time. The scheduled arrival time of the bus is:
Answer and solution
Correct answer: B
Solution
Answer: B — 5:00 pm
1. Let the scheduled journey take T hours and let the total distance be D km.
2. On the first day, the bus travels at 60 km per hour and arrives 3 hours late. Therefore,
3. D = 60(T + 3). ...(1)
4. On the second day, two-thirds of the distance is covered in one-third of the scheduled time. The remaining one-third is covered at 40 km per hour.
5. Time taken for the first part = T/3.
Time taken for the remaining part = D/(3 × 40) = D/120.
6. The bus arrives on time, so:
7. T/3 + D/120 = T
8. This gives D = 80T. ...(2)
9. Putting D = 80T in equation (1):
10. 80T = 60(T + 3)
80T = 60T + 180
20T = 180
T = 9 hours.
11. The bus starts at 8 am and the scheduled journey takes 9 hours.
12. Answer: 5:00 pm.
Why this method works:
The second-day remaining distance is D/3 and its available time is 2T/3. At 40 km/h this gives D/3=40(2T/3), hence D=80T. This relation is about the whole scheduled journey, not an average of the two speeds.
Independent check and common trap:
T=9 gives D=720 km. The first-day duration is 720/60=12 hours, exactly 3 hours late. On day two, 480 km takes 3 hours and the final 240 km takes 6 hours. Both descriptions agree; 8 am plus 9 hours is 5 pm.
Apply: Remaining one-third at 40 takes two-thirds of T: D=80T. Also D=60(T+3), so 20T=180 and T=9; arrival 5pm.
Check: T=9 gives D=720 km. The first-day duration is 720/60=12 hours, exactly 3 hours late. On day two, 480 km takes 3 hours and the final 240 km takes 6 hours. Both descriptions agree; 8 am plus 9 hours is 5 pm.
1. Let the scheduled journey take T hours and let the total distance be D km.
2. On the first day, the bus travels at 60 km per hour and arrives 3 hours late. Therefore,
3. D = 60(T + 3). ...(1)
4. On the second day, two-thirds of the distance is covered in one-third of the scheduled time. The remaining one-third is covered at 40 km per hour.
5. Time taken for the first part = T/3.
Time taken for the remaining part = D/(3 × 40) = D/120.
6. The bus arrives on time, so:
7. T/3 + D/120 = T
8. This gives D = 80T. ...(2)
9. Putting D = 80T in equation (1):
10. 80T = 60(T + 3)
80T = 60T + 180
20T = 180
T = 9 hours.
11. The bus starts at 8 am and the scheduled journey takes 9 hours.
12. Answer: 5:00 pm.
Why this method works:
The second-day remaining distance is D/3 and its available time is 2T/3. At 40 km/h this gives D/3=40(2T/3), hence D=80T. This relation is about the whole scheduled journey, not an average of the two speeds.
Independent check and common trap:
T=9 gives D=720 km. The first-day duration is 720/60=12 hours, exactly 3 hours late. On day two, 480 km takes 3 hours and the final 240 km takes 6 hours. Both descriptions agree; 8 am plus 9 hours is 5 pm.
G Strategy shortcut
Spot: Arithmetic – Speed, Time and Distance.Apply: Remaining one-third at 40 takes two-thirds of T: D=80T. Also D=60(T+3), so 20T=180 and T=9; arrival 5pm.
Check: T=9 gives D=720 km. The first-day duration is 720/60=12 hours, exactly 3 hours late. On day two, 480 km takes 3 hours and the final 240 km takes 6 hours. Both descriptions agree; 8 am plus 9 hours is 5 pm.
G Strategy: Spot: Arithmetic – Speed, Time and Distance. Apply: Remaining one-third at 40 takes two-thirds of T: D=80T. Also D=60(T+3), so 20T=180 and T=9; arrival 5pm. Check: T=9 gives D=720 km. The first-day duration is 720/60=12 hours, exactly 3 hours late. On day two, 480 km takes 3 hours and the final 240 km takes 6 hours. Both descriptions agree; 8 am plus 9 hours is 5 pm.