A) 12.4 hrs | B) 11.9 hrs |

C) 10.7 hrs | D) 9.9 hrs |

Explanation:

We know that

Time = Distance/speed

Required time = (10 1/4)/2 + (10 1/4)/6

= 41/8 + 41/6

= 287/24 = 11.9 hours.

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) 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.**

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.**

A) 222 m 44 cm | B) 204 m |

C) 201 m 21 cm | D) 208 m |

Explanation:

To find the minimum distance, we have to get the LCM of 75, 80, 85

Now, LCM of 75, 80, 85 = 5 x 15 x 16 x 17 = **20400**

Hence, the minimum distance each should walk so that thay can cover the distance in complete steps = 20400 cms = 20400/100 = **204 mts.**