A) 244 km | B) 144 km |

C) 444 km | D) 344 km |

Explanation:

Given a jeep travels a certain distance taking 6 hrs in forward journey

During the return journey, it takes 4 hrs with an increased speed 12 km/hr

Let 'x' be the distance.

Now, speed is given by distance/time.

Here the difference between both speeds = 12

$\frac{x}{4}-\frac{x}{6}=12$

$\frac{3x-2x}{12}=12$

=> x = 144 km

Therefore, the distance travelled by the jeep in forward or return journey is **x = 144 km.**

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.**

A) 60 | B) 120 |

C) 360 | D) 6 |

Explanation:

For this we have to find the LCM of 24, 36 and 30

LCM of 24, 36 and 30 = 360 sec

360/60 min = 6 minutes.

A) 5.75 kms | B) 7.36 kms |

C) 8.2 kms | D) 6.98 kms |

Explanation:

Total running distance in four weeks = (24 x 240) + (4 x 400)

= 5760 + 1600

= 7360 meters

= 7360/1000

=> 7.36 kms

A) 6:00 am | B) 6:30 pm |

C) 5:45 am | D) 5:52 pm |

Explanation:

Suppose they meet after 'h' hours

Then

3h + 4h = 17.5

7h = 17.5

h = 2.5 hours

So they meet at => 4 + 2.5 = 6:30 pm

A) 99 | B) 100 |

C) 89 | D) 1 |

Explanation:

Let, Distance between A and B = d

Distance travelled by P while it meets Q = d + 11

Distance travelled by Q while it meets P = d – 11

Distance travelled by Q while it meets R = d + 9

Distance travelled by R while it meets Q = d – 9

Here the ratio of speeds of P & Q => SP : SQ = d + 11 : d – 11

The ratio of speeds of Q & R => SQ : SR = d + 9 : d – 9

But given Ratio of speeds of P & R => P : R = 3 : 2

$\frac{SP}{SR}=\frac{SP}{SQ}x\frac{SQ}{SR}=\frac{\left(d+11\right){\displaystyle \left(d+9\right)}}{\left(d-11\right){\displaystyle \left(d-9\right)}}$

=> $\frac{\left(d+11\right)\left(d+9\right)}{\left(d-11\right)\left(d-9\right)}$ = 3/2

=> d = 1, 99

=> d = 99 satisfies.

Therefore, Distance between A and B = 99