Problem on Trains
- A train overtake two persons walking along a railway track. The first one walks at 4.5 km/h and the other one walks at 5.4 km/h. The train needs 8.4 s and 8.5 s respectively , to overtake them. What is the speed of the train, if both the persons are walking in the same direction as the train ?
-
View Hint View Answer Discuss in Forum
Speed of the person = 4 km/h = 4.5 x (5/18) = m/s = 5/4 m/s = 1.25 m/s
Speed of 2nd person = 5.4 km/h = 5.4 x (5/18) m/s = 3/2 m/s = 1.5 m/s
Let speed of train be x m/s.
Then, (x - 1.25) x 8.4 = (x - 1.5) x 8.5Correct Option: D
Speed of the person = 4 km/h = 4.5 x (5/18) = m/s = 5/4 m/s = 1.25 m/s
Speed of 2nd person = 5.4 km/h = 5.4 x (5/18) m/s = 3/2 m/s = 1.5 m/s
Let speed of train be x m/s.
Then, (x - 1.25) x 8.4 = (x - 1.5) x 8.5
⇒ 8.4x - 10.5 = 8.5x - 12.75
⇒ 0.1x = 2.25
⇒ x = 22.5
∴ Speed of the train = 225 x (18/5 ) = 81 km/h .
- Two trains A and B starts from Howrah and Patna towards Patna and Howrah respectively at the same time. After passing each other, they take 4, h, 48 min and 3 h, 20 min to reach Patna and Howrah, respectively. If the train from Howrah is moving at 45 km/h, then the speed of the other train is ?
-
View Hint View Answer Discuss in Forum
Given, a = 45 km/h, y = ?, t1 = 4 h 48 min and t2 = 3 h 20 min
Using y = a√t1 / t2= 45√4 h and 48 min /3 h and 20 minCorrect Option: D
Given, a = 45 km/h, y = ?, t1 = 4 h 48 min and t2 = 3 h 20 min
Using y = a√t1 / t2= 45√4 h and 48 min /3 h and 20 min
= 45√(24/5h) / (10/3 h) = 45√24 x 3/5 x 10
= 45 x √1.44 = 45 x 12 = 54 km/h
- A train travelling at 36 kmph completely crosses another train having half its length and travelling in the opposite direction at 54 kmph, in 12 second, If it also passes a railway platform in 11/2 minutes, the length of the platform is ?
-
View Hint View Answer Discuss in Forum
Let the length of slower train be L meters and the
length of faster train be (L/2) meters
Their relative speed = (36 + 54) km/hr
= 90 x (5/18)
= 25 m/sec
∵ 3L / (2 x 25) = 12
⇒ 3L = 600
⇒ L = 200
∴ Length of slower train = 200 meters
Let the length of platform by M meters
Then, 200 + M/[36 x (5/18)] = 90 sec.Correct Option: C
Let the length of slower train be L meters and the
length of faster train be (L/2) meters
Their relative speed = (36 + 54) km/hr
= 90 x (5/18)
= 25 m/sec
∵ 3L / (2 x 25) = 12
⇒ 3L = 600
⇒ L = 200
∴ Length of slower train = 200 meters
Let the length of platform by M meters
Then, 200 + M/[36 x (5/18)] = 90 sec.
⇒ 200 + M = 900
⇒ M = 700 meters
Length of platform = 700 meters.
