A) 8 min | B) 7.5 min |

C) 12.5 min | D) 12 min |

Explanation:

B runs around the track in 10 min.

i.e , Speed of B = 10 min per round

A beats B by 1 round

Time taken by A to complete 4 rounds

= Time taken by B to complete 3 rounds

= 30 min

A's speed = 30/4 min per round

= 7.5 min per round

Hence, if the race is only of one round A's time over the course = 7 min 30 sec

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.

A) 20 km | B) 16 km |

C) 14 km | D) 10 km |

Explanation:

Let distance = x km.

Time taken at 3 kmph : dist/speed = x/3 = 20 min late.

time taken at 4 kmph : x/4 = 30 min earlier

difference between time taken : 30-(-20) = 50 mins = 50/60 hours.

x/3- x/4 = 50/60

x/12 = 5/6

x = 10 km.