# Time and Distance Questions

**FACTS AND FORMULAE FOR TIME AND DISTANCE **

1.$Speed=\frac{Dis\mathrm{tan}ce}{Time},Time=\frac{Dis\mathrm{tan}ce}{Speed},Dis\mathrm{tan}ce=\left(Speed\times Time\right)$

2. x km/hr = $\left(x\times \frac{5}{18}\right)m/sec$

3. x m/sec = $\left(x\times \frac{18}{5}\right)km/hr$

4. If the ratio of the speeds of A and B is a:b , then the ratio of the times taken by them to cover the same distance is or b : a

5. Suppose a man covers a certain distance at x km/ hr and an equal distance at y km/hr . Then, the average speed during the whole journey is $\frac{2xy}{x+y}$km/hr.

A) 3.6 | B) 7.2 |

C) 8.4 | D) 10 |

Explanation:

Speed = $\left(\frac{600}{5\times 60}\right)m/sec$= 2 m/sec = 2 x (18/5) km/hr = 7.2 km/hr

A) 456 kms | B) 482 kms |

C) 552 kms | D) 556 kms |

Explanation:

Total distance travelled in 12 hours =(35+37+39+.....upto 12 terms)

This is an A.P with first term, a=35, number of terms,

n= 12,d=2.

Required distance = 12/2[2 x 35+{12-1) x 2]

=6(70+23)

= 552 kms.

A) 100 m | B) 150 m |

C) 190 m | D) 200 m |

Explanation:

Relative speed of the thief and policeman = (11 – 10) km/hr = 1 km/hr

Distance covered in 6 minutes = (1/60) x 6 km = 1/10 km = 100 m

Therefore, Distance between the thief and policeman = (200 – 100) m = 100 m.

A) 35.55 km/hr | B) 36 km/hr |

C) 71.11 km/hr | D) 71 km/hr |

Explanation:

Total time taken = (160/64 + 160/8)hrs

= 9/2 hrs.

Average speed = (320 x 2/9) km.hr

= 71.11 km/hr.

A) 90 km/h | B) 160 km/h |

C) 67.5 km/h | D) None of these |

Explanation:

$\frac{{s}_{1}}{{s}_{2}}=\sqrt{\frac{{t}_{2}}{{t}_{1}}}$

$\Rightarrow \frac{120}{{s}_{2}}=\sqrt{\frac{9}{16}}=\frac{3}{4}$

$\Rightarrow $${s}_{2}=160km/h$

A) 50 sec | B) 52 sec |

C) 54 sec | D) 56 sec |

Explanation:

Speed = 9 km/hr = 9 x (5/18) m/sec = 5/2 m/sec

Distance = (35 x 4) m = 140 m.

Time taken = 140 x (2/5) sec= 56 sec

A) 30 km/h | B) 15 km/h |

C) 25 km/h | D) 20 km/h |

Explanation:

If the speed of the faster horse be ${f}_{s}$ and that of slower horse be ${s}_{s}$ then

${f}_{s}+{s}_{s}=\frac{50}{1}=50$

and $\frac{50}{{s}_{s}}-\frac{50}{{f}_{s}}=\frac{5}{6}$

Now, you can go through options.

The speed of slower horse is 20km/h

Since, 20+30=50

and $\frac{50}{20}-\frac{50}{30}=\frac{5}{6}$

A) 28 km/h | B) 30 km/h |

C) 40 km/h | D) 20 km/h |

Explanation:

Let the normal speed be x km/h, then

$\frac{80}{x}-\frac{80}{(x+4)}=1$

$\Rightarrow $${x}^{2}+4x-320=0$

$\Rightarrow $x (x + 20) - 16 (x + 20) = 0

(x + 20 ) (x - 16) =0

x = 16 km/h

Therefore (x + 4) = 20 km/h

Therefore increased speed = 20 km/h