# 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) 69.5 km/hr | B) 70 km/hr |

C) 79 km/hr | D) 79.2 km/hr |

Explanation:

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

Then, x/y = 8 => x = 8y

Now, (x+264)/20 = y

8y + 264 = 20y

y = 22.

Speed = 22 m/sec = (22*18/5) km/hr = 79.2 km/hr.

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 = $\frac{3600}{18}xL$ m/hour

Speed of secxond train = $\frac{3600}{12}xL$ 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 = $\frac{2L}{500L}x3600$ sec

=14.4 sec

A) 80 kms | B) 60 kms |

C) 45 kms | D) 32 kms |

Explanation:

Let 'd' be the distance and 's' be the speed and 't' be the time

d=sxt

45 mins = 3/4 hr and 48 mins = 4/5 hr

As distance is same in both cases;

s(3/4) = (s-5)(4/5)

3s/4 = (4s-20)/5

15s = 16s-80

s = 80 km.

=> d = 80 x 3/4 = 60 kms.

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) 10 | B) 25 |

C) 12 | D) 20 |

Explanation:

Speed of train 1 = $\left(\frac{120}{10}\right)m/sec$ = 12 m/sec

Speed of train 2 =$\left(\frac{120}{15}\right)m/sec$ = 8 m/sec

if they travel in opposite direction, relative speed = 12 + 8 = 20 m/sec

distance covered = 120 + 120 = 240 m

time = distance/speed = 240/20 = 12 sec

A) 355 mts | B) 325 mts |

C) 365 mts | D) 312 mts |

Explanation:

Given speed = 63 km/hr = $63\times \frac{5}{18}=\frac{35}{2}$ m/s

Let the length of the bridge = x mts

Given time taken to cover the distance of (170 + x)mts is 30 sec.

We know speed = $\frac{dis\mathrm{tan}ce}{time}$m/s

$\Rightarrow \frac{35}{2}=\frac{170+x}{30}$

--> x = 355 mts.

A) 3 : 4 | B) 4 : 3 |

C) 2 : 3 | D) 3 : 2 |

Explanation:

Let us name the trains as A and B.

Then, (A's speed) : (B's speed)

= √b : √a = √16 : √9 = 4:3

A) 90 mts | B) 150 mts |

C) 120 mts | D) 100 mts |

Explanation:

We know that $Speed=\frac{dis\mathrm{tan}ce}{time}$

distance= speed * time

$\Rightarrow $ $d=\left(40\times \frac{5}{18}\right)\times 9$

d= 100 mts.