Directions (Q. Nos. 1-4) Study the following line graph to answer these questions.
Railway Time Schedule of an Express Train X Running Between City A and City H a → Arrival of train d→ Departure of train A, B, C, D, E, F, G and H are cities through which the train runs. a-d→ Indicates stoppage/halting of the train at the city station.
Question 1: The average speed of the train maintained between two successive stations was maximum between (a) E-F (b) F-G (c) G-H (d) Both G -H and F-G
Question 2: Between how many pairs of consecutive stations does the speed run below the overall average speed of the entire trip? (a) 4 (b) 1 (c) 3 (d) 2
Question 3: If the train stops at each city for 30% more time, then at what time will it reach the city H after departing from City A as per schedule? (a) Data insufficient (b) 17 : 03 (c) 16 : 41 (d) 16 : 58
Question 4: The train begins its onward journey from City A and it is extended to beyond City H to a City M due to some unavoidable reason. The train starts its return journey immediately after it reached City M. The train returns with a speed of 90 km/h without any stoppages in between and reaches City A at 2 : 25 am. Find the distance between City H and City M. (a) 40 km (b) 90 km (c) 70 km (d) None of these
Answers and Explanations
Answer 1: (b)
Therefore, the correct option is (b)
Answer 2: (c)
Overall average speed = 800/633 = 1.264 km/min
Hence, between stations B-C, D-E and G-H the speed of the train was less than the average speed of the entire journey. Correct option is (c)
Answer 3: (c) Now Increased stoppage time= 0.3 × 52 =15.6 min approximately 16 min
New arrival time = 16:25 + 16 = 16:41 min
Correct option is (c).
Answer 4: (d) Assuming the train starts its journey from M towards A at 16:25
Time taken to reach A from M= 24+2:25 – 16:25 = 10 hours.
Average speed = 90 Km/hr.
Total distance M-A = 90 × 10 = 900 Km.
Total distance A-H = 800 Km.
Therefore, Distance between H and M is 900-800 = 100 Km.