Boats And Streams - Aptitude MCQ Questions and Solutions with Explanations

Speed of downstream
D = 12 km/h
Speed of upstream
U = 8 km/h
Speed of boat in still water
$$\eqalign{ & = \frac{{D + U}}{2} \cr & = \frac{{20}}{2} \cr & = 10\,km/h \cr} $$
Time taken by the boat in still water
$$\eqalign{ & = \frac{{24\,km}}{{10\,km/hr}} \cr & = 2.4\,{\text{hours}} \cr} $$

$$\eqalign{ & {\text{Speed upstream}} \cr & {\text{ = }}\left( {\frac{{56}}{{1\frac{3}{4}}}} \right)km/hr \cr & = \left( {56 \times \frac{4}{7}} \right)km/hr \cr & = 32km/hr \cr & {\text{let speed downstream be }}x{\text{ km/hr}}{\text{.}} \cr & {\text{Then speed of boat in still water }} \cr & {\text{ = }}\frac{1}{2}\left( {x + 32} \right)km/hr \cr & \therefore {\text{ }}\frac{1}{2}\left( {x + 32} \right) = 36\,\,\, \Rightarrow x = 40 \cr & {\text{Hence , required time}} \cr & {\text{ = }}\left( {\frac{{50}}{{40}}} \right)hrs \cr & = 1\frac{2}{5}hrs \cr & = 1\,{\text{hour}}\,24\operatorname{minutes} \cr} $$

$$\eqalign{ & {\text{Let PQ = QR = }}x{\text{ }}km \cr & {\text{let speed downstream }} \cr & {\text{ = }}a{\text{ }}km/hr \cr & \,\,\,\,\,\,\,\,\,\, \to \,\,{\text{downstream}} \to \cr & {\text{P}}\overline {\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Q}}\,\,\,\,\,\,\,\,\,\,\,\,\,y\,\,\,\,\,\,\,\,\,\,\,} \,{\text{R}}\,\,\,\, \cr & {\text{and speed upstream }} \cr & {\text{ = }}b{\text{ }}km/hr{\text{ }} \cr & {\text{then, }}\frac{x}{a} + \frac{x}{b} = 10 \cr & \Rightarrow x = \frac{{10ab}}{{a + b}} \cr & {\text{and }}\frac{{2x}}{a} = 4 \cr & \Rightarrow x = \frac{{4a}}{2} = 2a \cr & {\text{from (i) and (ii) we have:}} \cr & 2a = \frac{{10ab}}{{a + b}} \cr & \Rightarrow 5b = a + b \cr & \Rightarrow a = 4b \cr & {\text{Required ratio }} \cr & {\text{ = }}\frac{{{\text{Speed in still water}}}}{{{\text{Speed of river}}}} \cr & = \frac{{\frac{1}{2}\left( {a + b} \right)}}{{\frac{1}{2}\left( {a - b} \right)}} \cr & = \frac{{\left( {a + b} \right)}}{{\left( {a - b} \right)}} \cr & = \frac{{4b + b}}{{4b - b}} \cr & = \frac{5}{3} \cr} $$

Rate downstream of boat
$$\eqalign{ & {\text{ = }}\left( {\frac{1}{{\frac{{15}}{{2 \times 60}}}}} \right)\,{\text{kmph}} \cr & = \frac{{2 \times 60}}{{15}}\,{\text{kmph}} \cr & = 8\,{\text{kmph}} \cr} $$
Rate downstream of boat = 5 kmph
Speed of boat in still water = $$\frac{1}{2}$$ (Rate downstream + Rate upstream)
$$\eqalign{ & = \frac{1}{2}\left( {8 + 5} \right) \cr & = \frac{{13}}{2} \cr & = 6\frac{1}{2}\,{\text{kmph}} \cr} $$

Let the speed of boat in still water = x km/hr,
and Speed of current = y km/hr
Rate downstream = (x + y) km/hr, and Rate upstream = (x – y) km/hr
Distance = Speed × Time
$$\eqalign{ & \therefore \left( {x - y} \right) \times 2t = \left( {x + y} \right) \times t \cr & \Rightarrow 2x - 2y = x + y \cr & \Rightarrow 2x - x = 2y + y \cr & \Rightarrow x = 3y \cr & \Rightarrow \frac{x}{y} = \frac{3}{1} = 3:1 \cr} $$

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Alternate Solution :
$$\eqalign{ & {\text{According to Question}} \cr & {\text{Downstream Speed}} = x + y \cr & {\text{Upstream Speed}} = x - y \cr & {\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & \therefore x + y = \frac{D}{T}\, . . . . .\,\left( {\text{i}} \right) \cr & \,\,\,\,x - y = \frac{D}{{2T}}\,. . . . .\,\left( {{\text{ii}}} \right) \cr & {\text{Solve equation}}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & x = \frac{{3D}}{{4T}},\,\,\,\,y = \frac{D}{{4T}} \cr & \therefore \frac{x}{y} = \frac{{3D}}{{4T}} \times \frac{{4T}}{D} \cr & \,\,\,\,\frac{x}{y} = \frac{3}{1} \cr & \,\,\,\,x:y = 3:1 \cr} $$

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

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

Let the speed of the stream be x km/hr
Then speed downstream = (10 + x) km/hr
Speed upstream
$$\eqalign{ & {\text{ = }}\left( {10 - x} \right)km/hr \cr & \therefore \frac{{26}}{{\left( {10 + x} \right)}} = \frac{{14}}{{\left( {10 - x} \right)}} \cr & \Rightarrow 260 - 26x = 140 + 14x \cr & \Rightarrow 40x = 120 \cr & \Rightarrow x = 3\,km/hr \cr} $$

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

S + W = 14 . . . . . . (i)
S - W = 8 . . . . . . . .(ii)
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from equation (i) & (ii)
S = 11 km/hr, W = 3 km/hr
Speed of current = 3 km/hr