# 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) 5 | B) 6 |

C) 7 | D) 10 |

Explanation:

Speed of train relative to man = (60 + 6) km/hr = 66 km/hr.

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

Time taken to pass the man = [110*(3/55)]m/sec = 6 sec.

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) 9 | B) 9.6 |

C) 10 | D) 10.8 |

Explanation:

Relative speed = (60 + 40) km/hr =[ 100 x ( 5 / 18 ) ]m/sec = ( 250 /9 ) m/sec.

Distance covered in crossing each other = (140 + 160) m = 300 m.

Required time = [ 300 x ( 9/250 ) ] sec = ( 54/ 5 )sec = 10.8 sec.

A) 50 m | B) 150 m |

C) 200 m | D) data inadequate |

Explanation:

Let the length of the train be x metres and its speed be y m/sec.

Then, (x/y)= 15 y =(x/15)

(x+100)/25 = x/15

=> 15(x + 100) = 25x

=> 15x + 1500 = 25x

=> 1500 = 10x

=> x = 150 m.

A) 10.8 | B) 18 |

C) 30 | D) 38.8 |

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) 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 sec | B) 11 sec |

C) 12 sec | D) 8 sec |

Explanation:

Relative Speed = 60 -40 = 20 x 5/18 = 100/18

Time = 50

Distance = 50 x 100/18 = 2500/9

Relative Speed = 60 + 40 = 100 x 5/18

Time = 2500/9 x 18/500 = 10 sec.