A) 5 kmph | B) 10 kmph |

C) 15 kmph | D) 45 kmph |

Explanation:

Speed of the boat downstream s=a/t= 60/3 = 20 kmph

Speed of the boat upstream s= d/t = 30/3= 10 kmph

Therefore, The speed of the stream = $\frac{speedofdownstream-speedofupstream}{2}$=5 kmph

A) 350 kms | B) 450 kms |

C) 900 kms | D) 700 kms |

Explanation:

Let the distance he covered each way = d kms

According to the question,

**d/45 - d/50 = 1**

**=> d = 450 kms.**

Hence, the total distance he covered in his way =** d + d = 2 d = 2 x 450 = 900 kms.**

A) 6 kmph | B) 8 kmph |

C) 9 kmph | D) 11 kmph |

Explanation:

Speed of boat in still water = 1/2 (12 + 6) = **9 kmph.**

A) 72 km | B) 36 km |

C) 56 km | D) 42 km |

Explanation:

Rate of her upstream = 12/2.5 = **4.8 km/hr**

Then, ATQ

Rate of downstream = 4.8 x 3 = **14.4 km/hr**

Hence, the distance she covers downstream in 5 hrs **= 14.4 x 5 = 72 kms.**

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) 2 kmph | B) 4 kmph |

C) 6 kmph | D) 8 kmph |

Explanation:

Let the distance in one direction = k kms

Speed in still water = 4.5 kmph

Speed of river = 1.5

Hence, speed in upstream = 4.5 - 1.5 = 3 kmph

Speed in downstream = 4.5 + 1.5 = 6 kmph

Time taken by Rajesh to row upwards = k/3 hrs

Time taken by Rajesh to row downwards = k/6 hrs

Now, required **Average speed** =$\frac{\mathbf{Total}\mathbf{}\mathbf{distance}}{\mathbf{Total}\mathbf{}\mathbf{speed}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{2}\mathbf{k}}{{\displaystyle \frac{\mathbf{k}}{\mathbf{3}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{k}}{\mathbf{6}}}}\mathbf{}\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{k}\mathrm{x}18}{6\mathrm{k}+3\mathrm{k}}=\mathbf{4}\mathbf{}\mathbf{kmph}$

Therefore, the average speed of the whole journey =** 4kmph.**

A) Only A and B | B) Only B and C |

C) All are required | D) Any one pair of A and B, B and C or C and A is sufficient |

Explanation:

Let distance between A & B = d km

Let speed in still water = x kmph

Let speed of current = y kmph

from the given data,

d/x = 2

From A) we get d

From B) we get d/x+y

From C) we get y

So, Any one pair of A and B, B and C or C and A is sufficient to give the answer i.e, the speed of upstream.

A) 5:1 | B) 3:1 |

C) 4:1 | D) 2:1 |

Explanation:

Let the speed of the man in still water = **p** kmph

Speed of the current = **s** kmph

Now, according to the questions

**(p + s) x 10 = (p - s) x 15**

2p + 2s = 3p - 3s

**=> p : s = 5 : 1**

Hence, ratio of his speed to that of current** = 5:1.**

A) 4 kmph | B) 6 kmph |

C) 3 kmph | D) 2 kmph |

Explanation:

Let the speed of current = **'C'** km/hr

Given the speed of boat in still water = 6 kmph

Let **'d'** kms be the distance it covers.

According to the given data,

Boat takes thrice as much time in going the same distance against the current than going with the current

i.e, $\frac{\mathbf{d}}{\mathbf{8}\mathbf{}\mathbf{-}\mathbf{}\mathbf{C}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{}\mathbf{\times}\mathbf{}\frac{\mathbf{d}}{\mathbf{8}\mathbf{}\mathbf{+}\mathbf{}\mathbf{C}}$

$\Rightarrow 24-3\mathrm{C}=8+\mathrm{C}\phantom{\rule{0ex}{0ex}}\Rightarrow 4\mathrm{C}=16\phantom{\rule{0ex}{0ex}}\mathbf{\Rightarrow}\mathbf{}\mathbf{C}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{}\mathbf{kmph}$

Hence, the speed of the current **C = 4 kmph.**