A) 10 a.m | B) 10.30 a.m |

C) 11 a.m | D) 11.30 a.m |

Explanation:

Suppose they meet x hrs after 8 a.m

then,

[Distance moved by first in x hrs] + [Distance moved by second in (x-1) hrs] = 330.

Therefore, 60x + 75(x-1) = 330.

=> x=3.

So,they meet at (8+3) i.e, 11a.m.

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.

A) 40 kms | B) 30 kms |

C) 46 kms | D) 32 kms |

Explanation:

Let the actual distance travelled be x km.

Then x/8=(x+20)/12

=> 12x = 8x + 160

=> 4x = 160

=> x = 40 km.