A) 16 km | B) 18 km |

C) 21 km | D) 25 km |

Explanation:

Let the distance covered be D km.

$\frac{D}{10+4}+\frac{D}{10-4}=5$

$\frac{D}{14}+\frac{D}{6}=5$

10D = 42 x 5 = 210

=> D = 21 km

A) 3 kms | B) 5 kms |

C) 6 kms | D) 4 kms |

Explanation:

Let the place be at a distance of 'd' kms

From the given data,

$\frac{\mathbf{d}}{\mathbf{5}\mathbf{}\mathbf{-}\mathbf{}\mathbf{1}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{d}}{\mathbf{5}\mathbf{}\mathbf{+}\mathbf{}\mathbf{1}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{75}}{\mathbf{60}}\mathbf{}\mathbf{kms}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{d}}{4}+\frac{\mathrm{d}}{6}=\frac{5}{4}\phantom{\rule{0ex}{0ex}}$

5d/12 = 5/4 =>** d = 3 kms.**

Hence, the place is **3 kms** far.

A) 14 kmph | B) 13 kmph |

C) 12 kmph | D) 11 kmph |

Explanation:

Given that, upstream distance = 7 kms

Upstream speed = 7/42 x 60 = 10 kms

Speed of the stream = 3 kmph

Let speed in still water = M kmph, then

Upstream speed = M - 3 = 10

=> M = 13 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.