# Problems on Trains Questions

**FACTS AND FORMULAE FOR PROBLEMS ON TRAINS**

**1. **a km/hr = [a x (5/18)] m/s.

**2. **a m/s = [a x (18/5)] km/hr.

**3.** Time taken by a train of length l metres to pass a pole or a standing man or a signal post is equal to the time taken by the train to cover l metres.

**4.** Time taken by a train of length 1 metres to pass a stationary object of length b metres is the time taken by the train to cover (1 + b) metres.

**5. **Suppose two trains or two bodies are moving in the same direction at u m/s and v m/s, where u > v, then their relatives speed = (u - v) m/s.

**6. **Suppose two trains or two bodies are moving in opposite directions at u m/s and v m/s, then their relative speed = (u + v) m/s.

**7. **If two trains of length a metres and b metres are moving in opposite directions at u m/s and v m/s, then time taken by the trains to cross each other = $\frac{(a+b)}{(u+v)}$sec.

**8. **If two trains of length a metres and b metres are moving in the same direction at u m/s and v m/s, then the time taken by the faster train to cross the slower train = $\frac{\left(a+b\right)}{\left(u-v\right)}$sec.

**9. **If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then (A's speed) : (B’s speed) = $\left(\sqrt{b}:\sqrt{a}\right)$

A) 68 kmph | B) 52 kmph |

C) 76 kmph | D) 50 kmph |

Explanation:

Let the speed of the faster train be 'X' kmph,

Then their relative speed= X - 48 kmph

To cross slower train by faster train,

Distance need to be cover = (400 + 600)m = 1 km. and

Time required = 180 sec = 180/3600 hr = 1/20 hr.

Time = Distance/Speed

=> 1/20 = 1/(x-48)

X = 68 kmph

A) 230 m | B) 240 m |

C) 260 m | D) 320 m |

Explanation:

Relative speed = (120 + 80) km/hr

=(200*5/18)m/s = (500/9)m/s

Let the length of the other train be x metres.

Then, x+270/9 = 500/9

=> x + 270 = 500

=> x = 230.

A) 60 sec | B) 48 sec |

C) 45 sec | D) 34 sec |

Explanation:

Relative speed = 42 + 36 = 78 km/hr = $\frac{65}{3}$ m/s

Distance = (520 + 520) =1040 mts.

Time = $1040\times \frac{3}{65}$= 48 sec

A) 387 kms | B) 242 kms |

C) 145 kms | D) 444 kms |

Explanation:

1h ----- 5 kms

? ------ 60 kms

Time = 12 hrs

Relative Speed = 16 + 21 = 37 kmph

T = 12 hrs

D = S x T = 37 x 12 = 444 kms.

A) 1 : 2 | B) 3 : 1 |

C) 4 : 7 | D) 3 : 2 |

Explanation:

Let the speeds of the two trains be x m/sec and y m/sec respectively.

Then, length of the first train = 27 x meters, and

length of the second train = 17 y meters.

(27 x + 17 y) / (x + y) = 23

=> 27 x + 17 y = 23 x + 23 y

=> 4 x = 6 y

=> x/y = 3/2.

A) 58 sec | B) 62 sec |

C) 60 sec | D) 50 sec |

Explanation:

The distance to be covered = Sum of their lengths = 200 + 300 = 500 m.

Relative speed = 72 -36 = 36 kmph = 36 x 5/18 = 10 mps.

Time required = d/s = 500/10 = 50 sec.

A) 10:10 pm | B) 9:50 pm |

C) 11:30 pm | D) 10:30 pm |

Explanation:

Distance = 70 x 1 ½ = 105 km

Relative Speed = 85 – 70 = 15

Time = 105/15 = 7 hrs

4:30 + 7 hrs = 11.30 p.m.

A) 60 mts | B) 120 mts |

C) 240 mts | D) 360 mts |

Explanation:

Speed of the Train K is given by s = d/t = 240/20 = 12 m/s

Distance covered by Train K in 50 seconds = 12 x 50 = 600 mts.

But it crosses Train L in 50 seconds

Therefore, the length of the Train L is = 600 - 240 = 360 mts.