A) 2h | B) 3h |

C) 3.5h | D) 4.25h |

Explanation:

Time ( when X was 30 km ahead of Y) = (120-30)/20 =4.5h

Time ( when Y was 30 km ahead of X) = (120+30)/20 = 7.5 h

Thus, required difference in time = 3h

A) 55 kms | B) 35 kms |

C) 25 kms | D) 65 kms |

Explanation:

Let x and y be the respective km's travelled by man via taxi and by his own car.

Given x + y = 90 => x = 90 - y

But according to the question,

7x + 6y = 595

7(90-y) + 6y = 595

=> 630 - 7y + 6y = 595

=> y = 630 - 595 = 35

=> x = 90 - 35 = 55

Therefore, the distance travelled by taxi is 55 kms.

A) 12 pm | B) 1 pm |

C) 11 am | D) 11:30 am |

Explanation:

let 't' be the time after which they met since L starts.

Given K is 50% faster than L

50 t + 1.5*50(t-1) = 300

50 t +75 t = 300 + 75

t = 375 / 125 = 3 hrs past the time that L starts

So they meet at (9 + 3)hrs = 12:00 noon.

A) 36 kmph | B) 32 kmph |

C) 28 kmph | D) 34 kmph |

Explanation:

Let the speed of goods truck be 'p' kmph

Distance covered by goods truck in 10hrs (4+6) = Distance covered by other truck in 4 hrs.

10p = 4 x 90

p = 36kmph.

Therefore, the speed of the goods truck is p = 36 kmph.

A) 39 days | B) 37 days |

C) 35 days | D) 33 days |

Explanation:

Distance d = 1200km

let S be the speed

he walks 15 hours a day(i.e 24 - 9)

so totally he walks for 70 x 15 = 1050hrs.

S = 1200/1050 => 120/105 = 24/21 => 8/7kmph

given 1 1/2 of previous speed

so 3/2 * 8/7= 24/14 = 12/7

New speed = 12/7kmph

Now he rests 10 hrs a day that means he walks 14 hrs a day.

time = 840 x 7 /12 => 490 hrs

=> 490/14 = 35 days

So he will take 35 days to cover 840 km.

A) 25 kmph | B) 24 kmph |

C) 26 kmph | D) 22 kmph |

Explanation:

Average speed=total distance/total time

Let distance to store be K

then, total time =(K/20)+(K/30)=K/12

and, total time =(2K)

so average speed= 2K / (K/12) = 24kmph.