Boats And Streams - Aptitude MCQ Questions and Solutions with Explanations

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

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

Speed upstream = 7.5 kmph
Speed downstream = 10.5 kmph
∴ Total time taken
= $$\left( {\frac{{105}}{{7.5}} + \frac{{105}}{{10.5}}} \right)$$   hours
= 24 hours

Let man's rate upstream be x kmph
Then, his rate downstream = 2x kmph
∴ (speed in still water) : (Speed of stream)
$$\eqalign{ & = {\frac{{2x + x}}{2}} : {\frac{{2x - x}}{2}} \cr & = \frac{{3x}}{2}:\frac{x}{2} \cr & = 3:1 \cr} $$

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

$$\eqalign{ & {\text{Speed of man in still water , }} \cr & x{\text{ }} = {\text{ }}\frac{9}{2}km/hr{\text{ }} \cr & {\text{let speed of stream = }}y{\text{ }}km/h \cr & {\text{Downstream speed = }} {\frac{9}{2} + y} \cr & {\text{Upstream speed = }} {\frac{9}{2} - y} \cr & {\text{Accroding to questions,}} \cr & {\text{2}} \times {\frac{{{\text{Distance}}}}{{ {\frac{9}{2} + y} }}} = \frac{{{\text{Distance}}}}{{ {\frac{9}{2} - y} }} \cr & \Rightarrow \frac{2}{{\frac{9}{2} + y}} = \frac{1}{{\frac{9}{2} - y}} \cr & \Rightarrow \frac{{2 \times 2}}{{9 + 2y}} = \frac{2}{{9 - 2y}} \cr & \Rightarrow \frac{2}{{9 + 2y}} = \frac{1}{{9 - 2y}} \cr & \Rightarrow 18 - 4y = 9 + 2y \cr & \Rightarrow 6y = 9 \cr & \Rightarrow y = \frac{9}{6} = \frac{3}{2} = 1.5\,km/h \cr} $$

let speed of boat in still water = x km/h
Speed of stream current = y km/h
According to question,
$$\eqalign{ & \frac{{24}}{{x - y}} + \frac{{36}}{{x + y}} = 6h\,......\,(i) \cr & \frac{{36}}{{x - y}} + \frac{{24}}{{x + y}} = \frac{{13}}{2}h\,......\,(ii) \cr} $$
In these type of questions, make factor of 24 and 36 and choose the common values which satisfy the above equations.
$$\eqalign{ & {\text{24 = 2,3,4,6,8,}}\boxed{12} \cr & 36 = 3,4,9,\boxed{12} \cr} $$
Choose the common factor i.e. Put this value in equation (i)
$$\eqalign{ & \frac{{24}}{{x - y}} + \frac{{36}}{{12}} = 6 \cr & \frac{{24}}{{x - y}} + 3 = 6 \cr & x - y = 8 \cr & \therefore x + y = 12 \cr & \therefore x = 10\,\,\,,\,\,\,\,y = 2 \cr & {\text{Speed of the current,}} \cr & y = 2{\text{ km/h}} \cr} $$

$$\eqalign{ & {\text{Time taken to cover 50 miles downstream}} \cr & {\text{ = }}\left( {\frac{{50}}{{60}}} \right)hr{\text{ = }}\frac{5}{6}hr.{\text{ }} \cr & {\text{Time taken to cover 50 miles upstream}} \cr & {\text{ = 1hr 15m = 1}}\frac{1}{4}hrs = \frac{5}{4}hrs \cr & {\text{Total time taken to cover 100 miles}} \cr & {\text{ = }}\left( {\frac{5}{6} + \frac{5}{4}} \right)hrs = \frac{{25}}{{12}}hrs \cr & \therefore {\text{Average speed }} \cr & {\text{ = }}\frac{{100}}{{\left( {\frac{{25}}{{12}}} \right)}}mph \cr & = \left( {\frac{{100 \times 12}}{{25}}} \right)mph \cr & = 48mph. \cr} $$

$$\eqalign{ & {\text{Upstream speed, U}} \cr & {\text{ = }}\frac{{24}}{6} = 4\,km/h \cr & {\text{Downstream speed , D}} \cr & {\text{ = }}\frac{{20}}{4} = 5\,km/h \cr & {\text{Speed of boat in still water , }}x \cr & = \frac{{{\text{D + U}}}}{2} = \frac{9}{2} = 4.5\,km/h \cr & {\text{Speed of water current, }}y \cr & = \frac{{{\text{D - U}}}}{2} = \frac{1}{2} = 0.5\,km/h \cr} $$

Speed of man in still water, x = 6 km/h
Speed of current, y = 2 km/h
Let distance = M
According to question
Upstream time = Downstream time + 4
$$\eqalign{ & \frac{M}{4} = \frac{M}{8} + 4 \cr & \frac{M}{4} = \frac{{M + 32}}{8} \cr & M = 32 \cr & \therefore {\text{Distance = 32 }}km \cr} $$