Average speed of train = 240/12 = 20 km/h
Average speed of bus = (3/4) x 20 = 3 x 5 = 15 km/h

Correct Option: C

Average speed of train = 240/12 = 20 km/h
Average speed of bus = (3/4) x 20 = 3 x 5 = 15 km/h
Required distance = 15 x 7 = 15 km/h

Let time taken to cover the distance = t
∴ Speed = (2t + 1)
⇒ t(2t + 1) = 1830 ⇒ 2t² + t = 1830

Correct Option: B

Let time taken to cover the distance = t
∴ Speed = (2t + 1)
⇒ t(2t + 1) = 1830 ⇒ 2t² + t = 1830
⇒ 2t² + t - 1830 = 0
∴ t = -1 ± √(1)² - 4 x 2 x(-1830) / 2 x 2
= -1 ± √1 + 14640 / 4
= -1 ± 121 / 4
Taking '+' sign t = -1 + 121 / 4 = 120/4 = 30
∴ Required ratio = (2 x 30 + 1) : 30 = 61 : 30

The distance covered by the train to cross the platform and bridge
= Length of the platform + Length of bridge + Distance between platform and bridge + Length of the train
= 165 + 135 + 30 + 110 = 440 m

Correct Option: A

The distance covered by the train to cross the platform and bridge
= Length of the platform + Length of bridge + Distance between platform and bridge + Length of the train
= 165 + 135 + 30 + 110 = 440 m
Speed of train = 110 /3 m/s
∴ Required time taken = 440/(110/3) = 12 s

Let the lengths of the trains A and B be a and 2a, respectively.
when a train crosses a pole, it covers the distance equal to its length.
∴ Required ratio of speeds = a / 25 : 2a /75 = 3 : 2

Correct Option: A

Let the lengths of the trains A and B be a and 2a, respectively.
when a train crosses a pole, it covers the distance equal to its length.
∴ Required ratio of speeds = a / 25 : 2a /75 = 3 : 2

Relative speed of the train = (40 - 22) km/hr = 18 km/hr
Relative speed of the train in m/sec = 18 x (5/18) = 5 m/sec.
Let the length of 2nd train be L meters.
Then, (125 + L) / 5 = 60 sec.

Correct Option: D

Relative speed of the train = (40 - 22) km/hr = 18 km/hr
Relative speed of the train in m/sec = 18 x (5/18) = 5 m/sec.
Let the length of 2nd train be L meters.
Then, (125 + L) / 5 = 60 sec.
⇒ 125 + L = 300
⇒ L = 175
∴ Length of second train = 175 meters