# 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) Option A | B) Option B |

C) Option C | D) Option D |

Explanation:

As Trains are moving in same direction,

Relative Speed = 60-20 = 40 kmph

--> $40\times \frac{5}{18}=\frac{100}{9}$ m/sec

Length of Train= Speed * Time

Length = $\frac{100}{9}\times 6$

$\Rightarrow \frac{200}{3}=66\frac{2}{3}mts\phantom{\rule{0ex}{0ex}}$

A) 15:07 hrs | B) 15:20 hrs |

C) 15:05 hrs | D) 15:00 hrs |

Explanation:

Next train comes at 17:00 hrs.

So, last train will be = 17:00hrs - 2:30hrs

= 14:30hrs

Announcement made after 37 min of the last train.

S0, 14:30hrs + 00:37 min

= 15:07 hrs.

A) 42 kmph | B) 72 kmph |

C) 36 kmph | D) 44 kmph |

Explanation:

Given speed of the first train = 60 km/hr = 60 x 5/18 = 50/3 m/s

Let the speed of the second train = x m/s

Then, **the difference in the speed** is given by

$\frac{\mathbf{50}}{\mathbf{3}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{120}}{\mathbf{18}}$

=> x = 10 m/s

=> 10 x 18/5 = **36 km/hr**

A) 21.4 and 42.8 | B) 25.2 and 50.4 |

C) 31.4208 and 62.8416 | D) 33.12 and 66.24 |

Explanation:

Birds speeds in mtrs/sec is 'x' and '2x' .

While flying from Engine to end, relative speed = (x+10) m/sec

from end to engine, flying speed = (2x - 10) mtr/sec

so

1000/(x+10) + 1000/(2x-10) = 187.5 secs

solving it, we get

so x = 8.728 m/sec and 2x= 17.456 m/sec

x = 31.4208 km/hr and 2x = 62.8416 km/hr.

A) 28 mts | B) 54 mts |

C) 24 mts | D) 50 mts |

Explanation:

Let the length of each train be x mts.

Then, distance covered = 2x mts.

Relative speed = 36 - 26 = 10 km/hr.

= 10 x 5/18 = 25/9 m/sec.

2x/36 = 25/9 => x = 50 mts.

A) 500 | B) 400 |

C) 360 | D) 480 |

Explanation:

Speed of train 1 = 48 kmph

Let the length of train 1 = 2x meter

Speed of train 2 = 42 kmph

Length of train 2 = x meter (because it is half of train 1's length)

Distance = 2x + x = 3x

Relative speed= 48+42 = 90 kmph = (90*5/18) m/s = 25 m/s

Time = 12 s

Distance/time = speed

=>3x/12 = 25 (25*12)/3

Length of the first train = 2x = 200 meter

Time taken to cross the platform= 45 s

Speed of train 1 = 48 kmph = 480/36 = 40/3 m/s

Distance = 200 + y [where y is the length of the platform]

x =100m => 200+y = 45* 40/3

y = 400m

A) 48 sec | B) 36 sec |

C) 18 sec | D) 72 sec |

Explanation:

Speed of the train relative to jogger = (45-9) km/hr = 36 km/hr =(36*5/18) m/sec = 10 m/sec

Distance to be covered =(240 + 120)m = 360 m

Time taken = 360/10 sec = 36 sec

A) 4/3 hrs | B) 2/3 hrs |

C) 3/2 hrs | D) 1/4 hrs |

Explanation:

Let the original speed and time is S and T

then distance = S x T

Now the speed changes to 2/3S and T is T+20

As the distance is same

S x T = 2/3Sx(T+20)

solving this we get t = 40 minutes

=40/60 = 2/3 hour