Boats & Streams - 02

Explanation:-

$$\eqalign{ & {\text{We}}\,{\text{can}}\,{\text{write}}\,{\text{three - quarters}}\,{\text{of}}\,{\text{akilometre}}\,{\text{as750}}\,{\text{metres}}. \cr & {\text{and}}\,11\frac{1}{4}\,{\text{minutes}}\,{\text{as}}\,{\text{675}}\,{\text{seconds}}{\text{.}} \cr & {\text{Rate}}\,{\text{upstream}} \cr & = \left( {\frac{{750}}{{675}}} \right)m/\sec = \frac{{10}}{9}m/\sec \cr & {\text{Rate}}\,{\text{downstream}} \cr & = \left( {\frac{{750}}{{450}}} \right)m/\sec = \frac{5}{3}m/\sec \cr & \therefore {\text{Rate}}\,{\text{in}}\,{\text{still}}\,{\text{water}} \cr & = \frac{1}{2}\left( {\frac{{10}}{9} + \frac{5}{3}} \right)m/\sec \cr & = \frac{{25}}{{18}}\,m/\sec \cr & = \left( {\frac{{25}}{{18}} \times \frac{{18}}{5}} \right)km/hr \cr & = 5\,km/hr \cr}$$

Explanation:-

$$\eqalign{ & {\text{Rate}}\,{\text{downstream}} \cr & = \left( {\frac{1}{{10}} \times 60} \right)km/hr = 6\,km/hr \cr & {\text{Rate}}\,{\text{upstream}} = 2\,km/hr \cr & {\text{Speed}}\,{\text{in}}\,{\text{still}}\,{\text{water}} \cr & = \frac{1}{2}\left( {6 + 2} \right)km/hr = 4\,km/hr \cr & \therefore {\text{Required}}\,{\text{time}} \cr & = \left( {\frac{5}{4}} \right)hrs \cr & = 1\frac{1}{4}hrs \cr & = 1\,hr\,15\,\min . \cr}$$

Explanation:-

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{boat}}\,{\text{in}}\,{\text{still}}\,{\text{water}}\,{\text{be}}\,x\,kmph{\text{,}}\,{\text{then}} \cr & {\text{Speed}}\,{\text{downstream}} \cr & = \left( {x + 3} \right)\,kmph \cr & {\text{Speed}}\,{\text{upstream}} \cr & = \left( {x - 3} \right)\,kmph \cr & \therefore \left( {x + 3} \right) \times 1 = \left( {x - 3} \right) \times \frac{3}{2} \cr & \Rightarrow 2x + 6 = 3x - 9 \cr & \Rightarrow x = 15\,kmph \cr}$$

Explanation:-

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{stream}}\,x\,mph.\,{\text{Then}}, \cr & {\text{Speed}}\,{\text{downstream}} = \left( {10 + x} \right)\,mph, \cr & {\text{Speed}}\,{\text{upstream}} = \left( {10 - x} \right)\,mph. \cr & \therefore \frac{{36}}{{\left( {10 - x} \right)}} - \frac{{36}}{{\left( {10 + x} \right)}} = \frac{{90}}{{60}} \cr & \Rightarrow 72x \times 60 = 90\left( {100 - {x^2}} \right) \cr & \Rightarrow {x^2} + 48x - 100 = 0 \cr & \Rightarrow \left( {x + 50} \right)\left( {x - 2} \right) = 0 \cr & \Rightarrow x = 2\,mph \cr}$$

Explanation:-

$$\eqalign{ & {\text{Speed}}\,{\text{downstream}} \cr & = \left( {15 + 3} \right)kmph = 18\,kmph. \cr & {\text{Distance}}\,{\text{travelled}} \cr & = \left( {18 \times \frac{{12}}{{60}}} \right)km = 3.6\,km \cr}$$

Explanation:-

$$\eqalign{ & {\text{Rate}}\,{\text{downstream}} \cr & = \left( {\frac{{16}}{2}} \right)kmph = 8\,kmph. \cr & {\text{Rate}}\,{\text{upstream}} \cr & = \left( {\frac{{16}}{4}} \right)kmph = 4\,kmph. \cr & \therefore {\text{Speed}}\,{\text{in}}\,{\text{still}}\,{\text{water}} \cr & = \frac{1}{2}\left( {8 + 4} \right)kmph = 6\,kmph \cr}$$

Explanation:-

$$\eqalign{ & {\text{Speed}}\,{\text{in}}\,{\text{stillwater}} \cr & = \frac{1}{2}\left( {11 + 5} \right)\,kmph \cr & = 8\,kmph \cr}$$

Explanation:-

Let the man's rate upstream be x kmph and that downstream be y kmph.
Then, distance covered upstream in 8 hrs 48 min = Distance covered downstream in 4 hrs.
$$\eqalign{ & \Rightarrow \left( {x \times 8\frac{4}{5}} \right) = \left( {y \times 4} \right) \cr & \Rightarrow \frac{{44}}{5}x = 4y \cr & \Rightarrow y = \frac{{11}}{5}x \cr & \therefore {\text{Required}}\,{\text{ratio}} \cr & = \left( {\frac{{y + x}}{2}} \right):\left( {\frac{{y - x}}{2}} \right) \cr & = \left( {\frac{{16x}}{5} \times \frac{1}{2}} \right):\left( {\frac{{6x}}{5} \times \frac{1}{2}} \right) \cr & = \frac{8}{5}:\frac{3}{5} \cr & = 8:3 \cr}$$

Explanation:-

Man's rate in still water
= (15 - 2.5) km/hr
= 12.5 km/hr.
Man's rate against the current
= (12.5 - 2.5) km/hr
= 10 km/hr.

Explanation:-

$$\eqalign{ & {\text{Speed}}\,{\text{downstream}} \cr & = \left( {13 + 4} \right)\,km/hr \cr & = 17\,km/hr \cr & {\text{Time}}\,{\text{taken}}\,{\text{to}}\,{\text{travel}}\,68\,km\,{\text{downstream}} \cr & = \frac{{68}}{{17}}\,hrs \cr & = 4\,hrs \cr}$$