# Boats and Streams Questions

A) 1.8h | B) 3h |

C) 4h | D) 5h |

Explanation:

Upstream speed = B-S

Downstream speed = B+s

B-S = 15/5 = 3 km/h

Again B= 4S

Therefore B-S = 3= 3S

=> S = 1 and B= 4 km/h

Therefore B+S = 5km/h

Therefore, Time during downstream = 15/5 = 3h

A) 180 km | B) 160 km |

C) 140 km | D) 120 km |

Explanation:

Speed in downstream = (14 + 4) km/hr = 18 km/hr;

Speed in upstream = (14 – 4) km/hr = 10 km/hr.

Let the distance between A and B be x km. Then,

x/18 + (x/2)/10 = 19 ⇔ x/18 + x/20 = 19 ⇒ x = 180 km.

A) 8 kmph | B) 6 kmph |

C) 7.5 kmph | D) 5.5 kmph |

Explanation:

Speed of the stream = 1

Motor boat speed in still water be = x kmph

Down Stream = x + 1 kmph

Up Stream = x - 1 kmph

[35/(x + 1)] + [35/(x - 1)] = 12

x = 6 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) 15 h | B) 16 h |

C) 8 h | D) 20 h |

Explanation:

If t1 and t2 are the upstream and down stream times. Then time taken in still water is given by

$\frac{2\times t1\times t2}{t1+t2}=\frac{2\times 12\times 24}{36}=16h$

A) 12 km/hr, 3 km/hr | B) 9 km/hr, 3 km/hr |

C) 8 km/hr, 2 km/hr | D) 9 km/hr, 6 km/hr |

Explanation:

Let the speed of the boat = p kmph

Let the speed of the river flow = q kmph

From the given data,

$\mathbf{2}\mathbf{}\mathbf{x}\mathbf{}\frac{\mathbf{28}}{\mathbf{p}\mathbf{}\mathbf{+}\mathbf{}\mathbf{q}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{28}}{\mathbf{p}\mathbf{}\mathbf{-}\mathbf{}\mathbf{q}}$

=> 56p - 56q -28p - 28q = 0

=> 28p = 84q

=> p = 3q.

Now, given that if

$\frac{\mathbf{28}}{\mathbf{3}\mathbf{q}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{q}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{28}}{\mathbf{3}\mathbf{q}\mathbf{}\mathbf{-}\mathbf{}\mathbf{2}\mathbf{q}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{672}}{\mathbf{60}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{28}{5\mathrm{q}}+\frac{28}{\mathrm{q}}=\frac{672}{60}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{}\mathbf{}\mathbf{q}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{}\mathbf{kmph}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{}\mathbf{}\mathbf{x}\mathbf{}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{q}\mathbf{}\mathbf{=}\mathbf{}\mathbf{9}\mathbf{}\mathbf{kmph}$

Hence, **the speed of the boat = p kmph = 9 kmph and the speed of the river flow = q kmph = 3 kmph.**

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