Relative speed  = (x + 50) km/hr  



Distance covered = (108 + 112) = 220 m.
∴  220  = 6  

⟹ 250 + 5x = 660
⟹ x = 82 km/hr.
Then, the length of the second train is  ❨  x  ❩  meters. 
2 
Relative speed = (48 + 42) kmph =  ❨  90 x  5  ❩  m/sec = 25 m/sec. 
18 
∴  [x + (x/2)]  = 12 or  3x  = 300 or x = 200. 
25  2 
∴ Length of first train = 200 m.
Let the length of platform be y meters.
Speed of the first train =  ❨  48 x  5  ❩  m/sec =  40  m/sec. 
18  3 
∴ (200 + y) x  3  = 45 
40 
⟹ 600 + 3y = 1800
⟹ y = 400 m.
Relative speed =  = (45 + 30) km/hr  



We have to find the time taken by the slower train to pass the DRIVER of the faster train and not the complete train.
So, distance covered = Length of the slower train.
Therefore, Distance covered = 500 m.
∴ Required time =  ❨  500 x  6  ❩  = 24 sec. 
125 
Then, length of the first train = 27x metres,
and length of the second train = 17y metres.
∴  27x + 17y  = 23 
x+ y 
⟹ 27x + 17y = 23x + 23y
⟹ 4x = 6y
⟹  x  =  3  . 
y  2 
2 kmph =  ❨  2 x  5  ❩  m/sec =  5  m/sec. 
18  9 
4 kmph =  ❨  4 x  5  ❩  m/sec =  10  m/sec. 
18  9 
Let the length of the train be x metres and its speed by y m/sec.
Then,  ❨  x  ❩  = 9 and  ❨  x  ❩  = 10. 


∴ 9y  5 = x and 10(9y  10) = 9x
⟹ 9y  x = 5 and 90y  9x = 100.
On solving, we get: x = 50.
∴ Length of the train is 50 m.
Then, speed of the faster train = 2x m/sec.
Relative speed = (x + 2x) m/sec = 3x m/sec.
∴  (100 + 100)  = 3x 
8 
⟹ 24x = 200
⟹ x =  25  . 
3 
So, speed of the faster train =  50  m/sec 
3 
=  ❨  50  x  18  ❩km/hr 
3  5 
= 60 km/hr.
Speed =  ❨  240  ❩m/sec = 10 m/sec. 
24 
∴ Required time =  ❨  240 + 650  ❩sec = 89 sec. 
10 
=  ❨  36 x  5  ❩m/sec 
18 
= 10 m/sec.
Distance to be covered = (240 + 120) m = 360 m.
Speed of the first train =  ❨  120  ❩  m/sec = 12 m/sec. 
10 
Speed of the second train =  ❨  120  ❩  m/sec = 8 m/sec. 
15 
Relative speed = (12 + 8) = 20 m/sec.
∴ Required time =  [  (120 + 120)  ]  sec = 12 sec. 
20 
=  ❨  66 x  5  ❩m/sec 
18 
=  ❨  55  ❩m/sec. 
3 
∴ Time taken to pass the man =  ❨  110 x  3  ❩sec = 6 sec. 
55 
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