Speed, Time and Distance
 Two trains of equal length take 10 seconds and 15 seconds respectively to cross a telegraph post. If the length of each train be 120 metres, in what time (in seconds) will they cross each other travelling in opposite direction ?

 16
 15
 12
 10

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When a train crosses a telegraph post, it covers its own length.
∴ Speed of first train = 120 = 12 m/sec. 10 Speed of second train = 120 = 8 m/sec. 15
Relative speed = 12 + 8 = 20 m/sec.Required time = Total length of trains Relative Speed = 2 × 120 = 12 seconds. 20 Correct Option: C
When a train crosses a telegraph post, it covers its own length.
∴ Speed of first train = 120 = 12 m/sec. 10 Speed of second train = 120 = 8 m/sec. 15
Relative speed = 12 + 8 = 20 m/sec.Required time = Total length of trains Relative Speed = 2 × 120 = 12 seconds. 20
 Two trains of length 70 m and 80 m are running at speed of 68 km/hr and 40 km/hr respectively on parallel tracks in opposite directions. In how many seconds will they pass each other ?

 10 sec
 8 sec
 5 sec
 3 sec

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Relative speed
= (68 + 40) kmph = 108 kmph= 108 × 5 m/s or 30 m/s 18 ∴ Required time = Sum of the lengths of both trains Relative Speed = 70 + 80 second = 5 seconds 30 Correct Option: C
Relative speed
= (68 + 40) kmph = 108 kmph= 108 × 5 m/s or 30 m/s 18 ∴ Required time = Sum of the lengths of both trains Relative Speed = 70 + 80 second = 5 seconds 30
 Two towns A and B are 500 km. apart. A train starts at 8 AM from A towards B at a speed of 70 km/hr. At 10 AM, another train starts from B towards A at a speed of 110 km/hr. When will the two trains meet ?

 1 PM
 12 Noon
 12.30 PM
 1.30 PM

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Let two trains meet after t hours when the train from town A leaves at 8 AM.
∴ Distance covered in t hours at 70 kmph + Distance covered in
(t – 2) hours at 110 kmph = 500km
∴ 70t + 110 (t – 2) = 500
⇒ 70t + 110t – 220 = 500
⇒ 180 t = 500 + 220 = 720⇒ t = 720 = 4 hours 180
Hence, the trains will meet at 12 noon.Correct Option: B
Let two trains meet after t hours when the train from town A leaves at 8 AM.
∴ Distance covered in t hours at 70 kmph + Distance covered in
(t – 2) hours at 110 kmph = 500km
∴ 70t + 110 (t – 2) = 500
⇒ 70t + 110t – 220 = 500
⇒ 180 t = 500 + 220 = 720⇒ t = 720 = 4 hours 180
Hence, the trains will meet at 12 noon.
 A train travelling at 48 km/hr crosses another train, having half its length and travelling in opposite direction at 42 km/hr, in 12 seconds. It also passes a railway platform in 45 seconds. The length of the railway platform is

 200 m
 300 m
 350 m
 400 m

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Let the length of the train travelling at 48 kmph be x metres.
Let the length of the platform be y metres.
Relative speed of train
= (48 + 42) kmph= 90 × 5 m./sec. 18
= 25 m./sec.
and 48 kmph= 48 × 5 = 40 m./sec. 18 3
According to the question,x + 1 = 2 25
= 12⇒ 3x = 12 2 × 25
⇒ 3x = 2 × 12 × 25 = 600
⇒ x = 200 m.Also, 200 + y = 45 40/3
⇒ 600 + 3y = 40 × 45
⇒ 3y = 1800 – 600 = 1200⇒ y = 1200 = 400 m. 3 Correct Option: D
Let the length of the train travelling at 48 kmph be x metres.
Let the length of the platform be y metres.
Relative speed of train
= (48 + 42) kmph= 90 × 5 m./sec. 18
= 25 m./sec.
and 48 kmph= 48 × 5 = 40 m./sec. 18 3
According to the question,x + 1 = 2 25
= 12⇒ 3x = 12 2 × 25
⇒ 3x = 2 × 12 × 25 = 600
⇒ x = 200 m.Also, 200 + y = 45 40/3
⇒ 600 + 3y = 40 × 45
⇒ 3y = 1800 – 600 = 1200⇒ y = 1200 = 400 m. 3
 A train, 150m long, passes a pole in 15 seconds and another train of the same length travelling in the opposite direction in 12 seconds. The speed of the second train is

 45 km./hr
 48 km./hr
 52 km./hr
 54 km./hr

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Let the speed of the second train be x m/s
Speed of first train = 150 = 10 m/sec 15
Relative speed of trains
= (x + 10) m/s
Total distance covered
= 150 + 150 = 300 metre∴ Time taken = 300 x + 10 ⇒ 300 = 12 x + 10
⇒ 12x + 120 = 300
⇒ 12x = 300 – 120 = 180⇒ x = 180 = 15 m/s 12 = 15 × 18 or 54 kmph. 5 Correct Option: D
Let the speed of the second train be x m/s
Speed of first train = 150 = 10 m/sec 15
Relative speed of trains
= (x + 10) m/s
Total distance covered
= 150 + 150 = 300 metre∴ Time taken = 300 x + 10 ⇒ 300 = 12 x + 10
⇒ 12x + 120 = 300
⇒ 12x = 300 – 120 = 180⇒ x = 180 = 15 m/s 12 = 15 × 18 or 54 kmph. 5