Boats & Streams - 01

Explanation:-

$$\eqalign{ & {\text{P - - - - - - - - - - - - }}\mathop {\text{ - }}\limits_{\text{Q}} {\text{ - - - - - - - - - - - - R}} \cr & {\text{ < - - - - - - - 16}}{\text{.40 - - - - - - - > }} \cr & \because {\text{Time required travelling from P to R}} \cr & {\text{ = 16H 40M}} \cr & \because {\text{PQ = QR}}\left( {\because {\text{Q is mid point}}} \right){\text{ }} \cr & {\text{Required time to travel PQ}} \cr & {\text{ = }}\frac{1}{2}\left( {{\text{16H 40M}}} \right) = \,\,{\text{8H 20M}} \cr & {\text{total required time from}} \cr & {\text{P to Q + Q to P = 12H}} \cr & {\text{ = 8H 20M + Q to P = 12H}} \cr & {\text{Time Q to P}} \cr & {\text{ = 12H - 8H 20M}} \cr & {\text{ = 3H 40M}} \cr & \because \overrightarrow {{\text{QP}}} {\text{ = }}\frac{1}{2}\overrightarrow {{\text{RP}}} {\text{ }}\,\,\left( {{\text{Distance}}} \right) \cr & \therefore {\text{Required time to }}\overrightarrow {{\text{RP}}} {\text{ }} \cr & {\text{ = 2}} \times \left[ {\overrightarrow {{\text{QP}}} } \right] \cr & {\text{ = 2}}\left[ {3{\text{H 40M}}} \right]{\text{ }} \cr & {\text{ = 7 H 20M}} \cr & {\text{ = 7}}\frac{1}{3}{\text{Hr}}{\text{. }} \cr}$$

Explanation:-

$$\eqalign{ & {\text{Speed of boat in still water}} \cr & {\text{ = }}\frac{{x + y}}{2} \cr & {\text{Where }}x{\text{ = downstream speed}} \cr & {\text{and }}y{\text{ = upstream speed}} \cr & {\text{Boat's speed}} \cr & {\text{ = }}\frac{{18 + 12}}{2} \cr & = \frac{{30}}{2} \cr & = 15\,km/hr \cr}$$

Explanation:-

$$\eqalign{ & {\text{Distance covered by a boat in 5 hours}} \cr & {\text{ = }}36{\text{ }}km \cr & {\text{Rate upstream of boat}} \cr & {\text{ = }}\,\frac{{36}}{5} = 7.2\,kmph \cr & {\text{Speed of the stream}}\, = 2.4\,kmph \cr & \therefore {\text{speed of the boat in stil water}} \cr & {\text{ = }}\left( {7.2 + 2.4} \right)kmph \cr & = 9.6\,\,kmph{\text{ }} \cr & \therefore {\text{Rate downstream of the boat}} \cr & {\text{ = }}\left( {9.6 + 2.4} \right)kmph \cr & = 12\,kmph \cr & \therefore {\text{Time taken in covering 78 }}km{\text{ distance}} \cr & {\text{ = }}\frac{{78}}{{12}} \cr & = 6.5\,hours. \cr}$$

Explanation:-

$$\eqalign{ & {\text{Speed of the boat in downstream }} \cr & {\text{ = }}\frac{{55}}{{2.5}} = \frac{{55 \times 10}}{{2.5}} = 22\,km/hr \cr & {\text{Then, speed of the boat in upstream}} \cr & {\text{ = }}\frac{{22}}{{2.2}} = \frac{{22 \times 10}}{{22}} = 10\,km/hr \cr & \therefore {\text{Speed of the boat in still water}} \cr & {\text{ = }}\frac{{22 + 10}}{2} = 16\,km/hr \cr}$$

Explanation:-

$$\eqalign{ & {\text{Relative speed }} \cr & {\text{ = 20 + 5 = 25 km/hr}} \cr & {\text{time = }}\frac{{100}}{{25}} = 4{\text{ hours}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{S + W = 14}}........{\text{(i)}} \cr & {\text{S }} - \,\,\,{\text{W = 8}}........{\text{(ii)}} \cr & {\text{ - - - - - - - - - - - - - - - - - - - - - - - - - }} \cr & {\text{from equation (i) \& (ii)}} \cr & {\text{S = 11 km/h}} \cr & {\text{W = 3 km/h}} \cr & {\text{Speed of current = 3 km/h}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{Suppose he moves 4 km downstream in }}x{\text{ hours}}{\text{.}} \cr & {\text{Then , Speed downstream}} \cr & {\text{ = }}\left( {\frac{4}{x}} \right)km/hr \cr & {\text{Speed upstream}} \cr & {\text{ = }}\left( {\frac{3}{x}} \right)km/hr \cr & \therefore \frac{{48}}{{\left( {\frac{3}{x}} \right)}} + \frac{{48}}{{\left( {\frac{4}{x}} \right)}} = 14\,\,\,or\,\,\,x = \frac{1}{2} \cr & So,\,{\text{Speed downstream}} \cr & {\text{ = 7 }}km/hr \cr & {\text{Speed uptream = 6 }}km/hr \cr & {\text{Rate of the stream}} \cr & {\text{ = }}\frac{1}{2}\left( {8 - 6} \right)km/hr \cr & = 1\,km/hr \cr}$$

Explanation:-

$$\eqalign{ & {\text{Given}} \cr & {\text{Speed of boat in downstream}} \cr & {\text{ = 15 km/h}} \cr & {\text{Speed of current = 3 km/h}} \cr & {\text{Speed of boat in still water}} \cr & {\text{ = 12 km/h}} \cr & {\text{Time taken at upstream }} \cr & {\text{ = }}\frac{{15}}{{12 - 3}} = \frac{{15}}{9}h = 1{\text{h }}\,40\min \cr & {\text{Time taken at downstream}} \cr & {\text{ = }}\frac{{15}}{{12 + 3}} = 1{\text{h}}\, \cr & {\text{total time }}\,{\text{ = 2 h 40 min }} \cr}$$

Explanation:-

$$\eqalign{ & {\text{According to question}} \cr & {\text{Speed of current }}y = {\text{ }}4{\text{ }}km/h \cr & {\text{Distance = }}12{\text{ }}km \cr & {\text{Speed in upstream }} \cr & {\text{ = }}\left( {x - y} \right)km/hr. \cr & {\text{Here }}x{\text{ is speed of boat in still water}} \cr & \,{\text{ = }}\frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & x - 4 = \frac{{12}}{5} \cr & 5x - 20 = 12 \cr & 5x = 32 \cr & x = 6.4\,km/hr \cr & {\text{Speed in downstream }} \cr & {\text{ = }}\left( {x + y} \right) = 6.4 + 4 \cr & = 10.4\,km/h \cr & \therefore {\text{Time = }}\frac{{{\text{Distance}}}}{{{\text{Speed }}}} \cr & {\text{Time = }}\frac{{15}}{{10.4}} = \frac{{150}}{{104}} \cr & = 1\,{\text{hour}}\,\,26\frac{7}{{13}}\,\,{\text{minutes}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{According to question}} \cr & {\text{Downstream speed = }}x + y \cr & {\text{Upstream speed = }}x - y \cr & {\text{Speed = }}\frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & \therefore x + y = \frac{{\text{D}}}{{\text{T}}}................(i) \cr & \,\,\,x + y = \frac{{\text{D}}}{{{\text{2T}}}}................(ii) \cr & {\text{Solve equation (i) and (ii) }} \cr & x{\text{ = }}\frac{{{\text{3D}}}}{{{\text{4T}}}},\,\,\,\,y = \frac{{\text{D}}}{{{\text{4T}}}} \cr & \therefore \frac{x}{y}\,\, = \frac{{{\text{3D}}}}{{{\text{4T}}}} \times \frac{{4{\text{T}}}}{{\text{D}}} = \frac{3}{1} = 3:1 \cr & {\text{Downstream (D)}}\,\,\,\,\,\,\,\,\,{\text{Upstream (U)}} \cr & {\text{T}} \to \,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \cr & {\text{S}} \to \,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \cr & {{\text{S}}_{\text{B}}} = \frac{{{{\text{S}}_{\text{D}}}{\text{ + }}{{\text{S}}_{\text{U}}}}}{2} = \frac{{2 + 1}}{2} = \frac{3}{2} \cr & {{\text{S}}_{\text{C}}} = \frac{{{{\text{S}}_{\text{D}}} - {{\text{S}}_{\text{U}}}}}{2} = \frac{{2 - 1}}{2} = \frac{1}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\text{S}}_{\text{B}}}\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,\,{{\text{S}}_{\text{C}}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \,\,\,\frac{3}{2}\,\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,\,\frac{1}{2} \cr & \,\,\,\,\,\,\,\,\,\,\, = \,\,\,\,3\,\,\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,\,\,1 \cr}$$