A) 120m | B) 150m |

C) 125m | D) None of these |

Explanation:

Let the length of the train e x meter, and let the speed of train be y km/h, then

and

From eq (1) and (2), we get

y = 12 km/h

x= 150 m

A) 54 sec | B) 36 sec |

C) 72 sec | D) 11 sec |

Explanation:

Time taken to meet the first time = length of track/relative speed

Given the length of the track is 600 m

The relative speed = 25 + 35 = 60 kmph = 60 x 5/18 m

Therefore Time = = 600/60 x (18/5) = 36 sec.

A) 1.8 kmph | B) 2 kmph |

C) 2.2 kmph | D) 1.5 kmph |

Explanation:

Speed of Boy is B = 4.5 kmph

Let the speed of the stream is S = x kmph

Then speed in Down Stream = 4.5 + x

speed in Up Stream = 4.5 - x

As the distance is same,

=> 4.5 + x = (4.5 - x)2

=> 4.5 + x = 9 -2x

3x = 4.5

x = 1.5 kmph

A) 4.58 kms | B) 6.35 kms |

C) 5.76 kms | D) 5.24 kms |

Explanation:

Speed in still water = 6 kmph

Stream speed = 1.2 kmph

Down stream = 7.2 kmph

Up Stream = 4.8 kmph

x/7.2 + x/4.8 = 1

x = 2.88

Total Distance = 2.88 x 2 = 5.76 kms

A) 55 kmph | B) 58 kmph |

C) 60 kmph | D) 66 kmph |

Explanation:

Speed on return trip = 150% of 50 = 75 km/hr.

Average speed = (2 x 50 x 75)/(50 + 75) = 60 km/hr.

A) 2/3 hrs | B) 3/4 hrs |

C) 1/3 hrs | D) 1/4 hrs |

Explanation:

Walking at 3/4th of usual rate implies that time taken would be 4/3th of the usual time. In other words, the time taken is 1/3rd more than his usual time

so 1/3rd of the usual time = 15min

or usual time = 3 x 15 = 45min = 45/60 hrs = 3/4 hrs.