# 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) 60 | B) 62 |

C) 64 | D) 65 |

Explanation:

Relative speed =280/9 m / sec = (280/9*18/5) kmph = 112 kmph.

Speed of goods train = (112 - 50) kmph = 62 kmph.

A) 10 | B) 8 |

C) 6 | D) 4 |

Explanation:

Distance = 110 m

Relative speed = 60 + 6 = 66 kmph (Since both the train and the man are in moving in opposite direction)

= (66*5/18) m/sec = 55/3 m/sec

Time taken to pass the man = (100*3/55) = 6 s

A) 42 | B) 36 |

C) 28 | D) 20 |

Explanation:

Distance covered = 120+120 = 240 m

Time = 12 s

Let the speed of each train = v. Then relative speed = v + v = 2v

2v = distance/time = 240/12 = 20 m/s

Speed of each train = v = 20/2 = 10 m/s

= 10 × 36/10 km/hr = 36 km/hr

A) 9.5 km/hr | B) 10 km/hr |

C) 10.5 km/hr | D) 11 km/hr |

Explanation:

Man's rate in still water = (15 - 2.5) km/hr = 12.5 km/hr.

Therefore, Man's rate against the current = (12.5 - 2.5) = **10 km/hr.**

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

C) 4:3 | D) 5:4 |

Explanation:

Note : 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) = (b : a)

Therefore, Ratio of the speeds of two trains = $\sqrt{9}:\sqrt{4}$ = 3 : 2

A) 2.5 min | B) 3 min |

C) 3.2 min | D) 3.5 min |

Explanation:

Total distance covered =$\left(\frac{7}{2}+\frac{1}{4}\right)$miles

=$\frac{15}{4}$miles

Time taken = $\left(\frac{15}{4*75}\right)$hrs

=$\frac{1}{20}$ hrs

=$\left(\frac{1}{20}*60\right)min$

= 3 min

A) 48 | B) 24 |

C) 38 | D) 36 |

Explanation:

Less Cogs more turns and less time less turns

$CogsTimeTurns\phantom{\rule{0ex}{0ex}}A544580\phantom{\rule{0ex}{0ex}}B328?$

Number of turns required=80 × 54/32 × 8/45 = 24 times

A) 10.30 | B) 10 |

C) 8.45 | D) 9.30 |

Explanation:

Assume both trains meet after x hours after 7 am

Distance covered by train starting from P in x hours = 20x km

Distance covered by train starting from Q in (x-1) hours = 25(x-1)

Total distance = 110

=> 20x + 25(x-1) = 110

=> 45x = 135

=> x= 3

Means, they meet after 3 hours after 7 am, ie, they meet at 10 am