A) 14.4 sec | B) 15.5 sec |

C) 18.8 sec | D) 20.2 sec |

Explanation:

Let the length of the train be L metres

Speed of first train = m/hour

Speed of secxond train = m/hour

When running in opposite directions, relative speed = 200 L + 300 L m/hour

Distance to be covered = L + L = 2L metre

Time taken = sec

=14.4 sec

A) 240 mts | B) 270 mts |

C) 260 mts | D) 250 mts |

Explanation:

Relative speed = 120 + 80 = 200 km/hr.

= 200 x 5/18 = 500/9 m/sec.

Let the length of the other train be L mts.

Then, (L + 260)/9 = 500/9 => L = 240 mts.

A) 49 kmph | B) 50 kmph |

C) 51 kmph | D) 52 kmph |

Explanation:

Speed of the train relative to man

= (125/10) m/sec = (25/2) m/sec.

[(25/2) x (18/5)] km/hr = 45 km/hr.

Let the speed of the train be 'x' km/hr.

Then, relative speed = (x - 4) km/hr.

x - 4 = 45 => x = 49 km/hr.

A) 42 sec | B) 44 sec |

C) 46 sec | D) 48 sec |

Explanation:

Relative speed = 60 + 90 = 150 km/hr.

= 150 x 5/18 = 125/3 m/sec.

Distance covered = 1.10 + 0.9 = 2 km = 2000 m.

Required time = 2000 x 3/125 = 48 sec.

A) 170 m | B) 100 m |

C) 270 m | D) 320 m |

Explanation:

Relative speed = (72 - 36) x 5/18 = 2 x 5 = 10 mps.

Distance covered in 32 sec = 32 x 10 = 320 m.

The length of the faster train = 320 m.

A) 360 m | B) 420 m |

C) 460 m | D) 320 m |

Explanation:

Let the length of the stationary train Y be LY

Given that length of train X, LX = 300 m

Let the speed of Train X be V.

Since the train X crosses train Y and a pole in 60 seconds and 25 seconds respectively.

=> 300/V = 25 ---> ( 1 )

(300 + LY) / V = 60 ---> ( 2 )

From (1) V = 300/25 = 12 m/sec.

From (2) (300 + LY)/12 = 60

=> 300 + LY = 60 (12) = 720

=> LY = 720 - 300 = 420 m

Length of the stationary train = 420 m