A) 360 kms | B) 480 kms |

C) 520 kms | D) 240 kms |

Explanation:

Now, the distance covered by Chennai express in 2 hrs = 60 x 2 = 120 kms

Let the Charminar Express takes **'****t' hrs** to catch Chennai express

=> 80 x t = 60 x (2 + t)

=> 80 t = 120 + 60t

=> **t = 6 hrs**

Therefore, the distance away from Hyderabad the two trains meet =** 80 x 6 = 480 kms.**

**1. How to find Speed(s) if distance(d) & time(t) is given:**

$\mathbf{Speed}\mathbf{}\mathbf{\left(}\mathbf{S}\mathbf{\right)}\mathbf{}\mathbf{=}\frac{\mathbf{}\mathbf{Distance}\mathbf{}\mathbf{\left(}\mathbf{D}\mathbf{\right)}}{\mathbf{Time}\mathbf{}\mathbf{\left(}\mathbf{T}\mathbf{\right)}}\mathbf{}$

**Ex:** Find speed if a person travels 4 kms in 2 hrs?

**Speed = D/T = 4/2 = 2 kmph.**

**2. ** Similarly, we can find distance (d) if speed (s) & time (t) is given by

**Distance (D) = Speed (S) x Time (T)**

**Ex :** Find distance if a person with a speed of 2 kmph in 2 hrs?

**Distance D = S X T = 2 x 2 = 4 kms. **

**3. **Similarly, we can find time (t) if speed (s) & distance (d) is given by

**Time (T) = **$\frac{\mathbf{Distance}\mathbf{}\mathbf{\left(}\mathbf{D}\mathbf{\right)}}{\mathbf{Speed}\mathbf{}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}$

**Ex :** Find in what time a person travels 4 kms with a speed of 2 kmph?

**Time T = D/S = 4/2 = 2 hrs.**

**4. **How to convert** km/hr into m/sec :**

**$\mathbf{P}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{hr}\mathbf{}\mathbf{=}\mathbf{}\left(\mathbf{P}\mathbf{}\mathbf{x}\mathbf{}\frac{\mathbf{5}}{\mathbf{18}}\right)\mathbf{}\mathbf{m}\mathbf{/}\mathbf{sec}$**

**Ex : **Convert 36 kmph into m/sec?

**36 kmph = 36 x 5/18 = 10 m/sec **

**5. **How to convert** m/sec into km/hr :**

**$\mathbf{P}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{sec}\mathbf{}\mathbf{=}\mathbf{}\left(\mathbf{P}\mathbf{}\mathbf{x}\mathbf{}\frac{\mathbf{18}}{\mathbf{5}}\right)\mathbf{}\mathbf{kmph}$.**

**Ex : **Convert 10 m/sec into km/hr?

**10 m/sec = 10 x 18/5 = 36 kmph.**

**6. **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 **b : a.**

**7.** 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 $\frac{\mathbf{2}\mathbf{xy}}{\mathbf{x}\mathbf{+}\mathbf{y}}$ **km/hr.**

A) 105 mts | B) 115 mts |

C) 120 mts | D) 140 mts |

Explanation:

Speeds of two trains = 30 kmph and 58 kmph

=> Relative speed = 58 - 30 = 28 kmph = 28 x 5/18 m/s = 70/9 m/s

Given a man takes time to cross length of faster train = 18 sec

Now, required Length of faster train = speed x time = **70/9 x 18 = 140 mts.**

A) 200 mts | B) 180 mts |

C) 160 mts | D) 145 mts |

Explanation:

Length of train A = 48 x 9 x 5/18 = 120 mts

Length of train B = 48 x 24 x 5/18 - 120

=> 320 - 120 = 200 mts.

A) 132 km | B) 264 km |

C) 134 km | D) 236 km |

Explanation:

Let the distance travelled by Tilak in first case or second case = d kms

Now, from the given data,

d/20 = d/22 + 36 min

=> d/20 = d/22 + 3/5 hrs

=> d = 132 km.

Hence, the total distance travelled by him = d + d = 132 + 132 = 264 kms.

A) 15 kms | B) 9 kms |

C) 6 kms | D) 18 kms |

Explanation:

The first train covers 180 kms in 5 hrs

=> Speed = 180/5 = 36 kmph

Now the second train covers the same distance in 1 hour less than the first train => 4 hrs

=> Speed of the second train = 180/4 = 45 kmph

Now, required difference in distance in 1 hour = 45 - 36 = 9 kms.

A) 32 kmph | B) 20 kmph |

C) 18 kmph | D) 24 kmph |

Explanation:

Let the total distance of the journey of a man = d kms

Now, the average speed of the entire journey = $\frac{\mathbf{d}}{{\displaystyle \frac{\mathbf{d}}{\mathbf{120}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{d}}{\mathbf{120}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{d}}{\mathbf{40}}}}\mathbf{}$ = **24 kmph.**

A) 60 kmph | B) 62 kmph |

C) 64 kmph | D) 63 kmph |

Explanation:

Speed of lorry = $\frac{\mathbf{360}}{\mathbf{12}}$ = 30 kmph

Speed of van =$\frac{\mathbf{250}}{\mathbf{100}}\mathbf{}\mathbf{x}\mathbf{}\mathbf{30}$ = 75 kmph

Speed of bike = $\frac{\mathbf{3}}{\mathbf{5}}\mathbf{}\mathbf{x}\mathbf{}\mathbf{75}$ = 45 kmph

Therefore, now required average speed of bike and van = $\frac{\mathbf{75}\mathbf{}\mathbf{+}\mathbf{}\mathbf{45}}{\mathbf{2}}$= **60 kmph.**

A) 14 kmph | B) 13 kmph |

C) 12 kmph | D) 11 kmph |

Explanation:

The distance is constant in this case.

Let the time taken for travel with a speed of 10 kmph be '**t**'.

Now the speed of 15 kmph is **3/2** times the speed of 10 kmph.

Therefore, time taken with the speed of 15 kmph will be 2t/3 (**speed is inversely proportional to time**)

Extra time taken = t - 2t/3 = t/3

=> 1pm - 11am = 2hrs

=> t/3 = 2h

=> t = 6 hrs.

Now, Distance = **speed x time** = 10 x 6 = 60 kms

Time he takes to reach at noon = 6 - 1 = 5 hrs

Now, Speed = 60/5 = **12 kmph.**