A) 2 9/7 km | B) 3 7/5 km |

C) 1 3/4 km | D) 4 3/8 km |

Explanation:

Relative Speed = 5 ¾ - 4 ½ = 1 ¼

Time = 3 ½ h.

Distance = 5/4 x 7/2 = 35/8 = 4 3/8 km.

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

A) 36 sec | B) 39 sec |

C) 41 sec | D) 43 sec |

Explanation:

Speed = 44 kmph x 5/18 = 110/9 m/s

We know that, Time = distance/speed

Time = (360 + 140) / (110/9)

= 500 x 9/110 = 41 sec.

A) 41.25 days | B) 37.5 days |

C) 39.75 days | D) 40 days |

Explanation:

Given Karthik can cover the distance of 200 kms resting 2 hrs per day in 22 days.

Let the initial speed be '1'

Hence, Time = 22(24hrs - 2)

Noe new speed = 2/3

Let the number of days he take be '**D**'

Therefore, $\frac{\mathbf{22}\left(\mathbf{24}\mathbf{-}\mathbf{2}\right)\mathbf{x}\mathbf{3}}{\mathbf{200}}\mathbf{=}\frac{\mathbf{D}\left(\mathbf{24}\mathbf{-}\mathbf{2}\right)\mathbf{x}\mathbf{2}}{\mathbf{250}}$

D = 11 x 3 x 5/4 = **41.25 days**. = 41 1/4 days.