Boats And Streams — Part 1 MCQs with Explanations

Speed downstream = (13 + 4) km/hr = 17 km/hr Time taken to travel 68 km downstream = $$\frac{{68}}{{17}}$$ hours = 4 hours

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.

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. $$\eqalign{ & \Rightarrow {x \times 8\frac{4}{5}} = {y \times 4} \cr & \Rightarrow \frac{{44}}{5}x = 4y \cr & \Rightarrow y = \frac{{11}}{5}x \cr & \therefore {\text{Required}}\,{\text{ratio}} \cr & = {\frac{{y + x}}{2}} : {\frac{{y - x}}{2}} \cr & = \left( {\frac{{16x}}{5} \times \frac{1}{2}} \right):\left( {\frac{{6x}}{5} \times \frac{1}{2}} \right) \cr & = \frac{8}{5}:\frac{3}{5} \cr & = 8:3 \cr} $$

Let the speed of the stream be x km/hr Then, Speed downstream = (15 + x) km/hr Speed upstream = (15 - x) km/hr $$\eqalign{ & \therefore \frac{{30}}{{ {15 + x} }} + \frac{{30}}{{ {15 - x} }} = 4\frac{1}{2} \cr & \Rightarrow \frac{{900}}{{225 - {x^2}}} = \frac{9}{2} \cr & \Rightarrow 9{x^2} = 225 \cr & \Rightarrow {x^2} = 25 \cr & \Rightarrow x = 5\,km/hr \cr} $$

Speed in still water = $$\frac{1}{2}$$(11 + 5) kmph = 8 kmph

Rate downstream = $$\frac{{16}}{2}$$ kmph = 8 kmph Rate upstream = $$\frac{{16}}{4}$$ kmph = 4 kmph ∴ Speed in still water = $$\frac{1}{2}$$(8 + 4) kmph = 6 kmph

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

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

Speed downstream = (5 + 1) kmph = 6 kmph Speed upstream = (5 - 1) kmph = 4 kmph Let the required distance be x km Then, $$\frac{x}{6} + \frac{x}{4}$$ = 1 ⇒ 2x + 3x = 12 ⇒ 5x = 12 ⇒ x = 2.4 km

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

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

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 $$\eqalign{ & \frac{M}{4} = \frac{M}{8} + 4 \cr & \frac{M}{4} = \frac{{M + 32}}{8} \cr & M = 32 \cr & \therefore {\text{Distance = 32 }}km \cr} $$

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 $$\eqalign{ & \therefore \left( {x - y} \right) \times 2t = \left( {x + y} \right) \times t \cr & \Rightarrow 2x - 2y = x + y \cr & \Rightarrow 2x - x = 2y + y \cr & \Rightarrow x = 3y \cr & \Rightarrow \frac{x}{y} = \frac{3}{1} = 3:1 \cr} $$ 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) _________________ from equation (i) & (ii) S = 11 km/hr, W = 3 km/hr Speed of current = 3 km/hr

Relative speed = 20 + 5 = 25 km/hr Time = $$\frac{{100}}{{25}}$$ = 4 hours

$$\eqalign{ & {\text{Speed of the boat in downstream }} \cr & {\text{ = }}\frac{{55}}{{2.5}} = \frac{{55 \times 10}}{{25}} = 22\,km/hr \cr & {\text{Then, speed of the boat in upstream}} \cr & {\text{ = }}\frac{{22}}{{2.2}} = \frac{{22 \times 10}}{{22}} = 10\,km/hr \cr & \therefore {\text{Speed of the boat in still water}} \cr & {\text{ = }}\frac{{22 + 10}}{2} = 16\,km/hr \cr} $$

Let the speed of the current be x kmph Then speed of the boat in still water = 5x $$\eqalign{ & \therefore {\text{Downstream speed}} \cr & {\text{ = }}\left( {5x + x} \right) = 6x\,kmph \cr & {\text{Upstream speed}} \cr & {\text{ = }}\left( {5x - x} \right) = 4x\,kmph \cr & {\text{Now, }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{13}}{\text{.2km}}\,\,\,\, \cr & {\text{A}}\overline {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} {\text{B}} \cr & {\text{According to question,}} \cr & {\text{1}}{\text{.1}} \times {\text{6}}x = 13.2 \cr & \Rightarrow 6.6x = 13.2 \cr & \Rightarrow x = \frac{{13.2}}{{6.6}} \cr & \therefore x = 2\,kmph \cr & \therefore {\text{Upstream speed}} \cr & {\text{ = 4}}x = 4 \times 2 = 8\,kmph \cr & \therefore {\text{312 minutes}}\, \cr & = 5\frac{1}{5}\,hours \cr & = \frac{{26}}{5}\,hours \cr & \therefore {\text{Required distance travelled upstream}} \cr & {\text{ = Speed }} \times {\text{Time}} \cr & {\text{ = 8}} \times \frac{{26}}{5} = 41.6\,km \cr} $$

Distance covered by a boat in 5 hours = 36 km Rate upstream of boat = $$\frac{{36}}{5}$$ = 7.2 kmph Speed of the stream = 2.4 kmph ∴ Speed of the boat in still water = (7.2 + 2.4) kmph = 9.6 kmph ∴ Rate downstream of the boat = (9.6 + 2.4) kmph = 12 kmph ∴ Time taken in covering 78 km distance = $$\frac{{78}}{{12}}$$ = 6.5 hours

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

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

Speed of boat in upstream = $$\frac{{35}}{7}$$ = 5 km/hr Speed of boat in downstream = $$\frac{{35}}{5}$$ = 7 km/hr Let the speed of boat x and speed of stream be y Now, x - y = 5 x + y = 7 $$\overline {\,2{\text{x}} = 12\,} $$ ∴ x = 6 km/hr

$$\eqalign{ & {\text{Let, speed of boat}} = x \cr & {\text{Speed of stream}} = y \cr & \Rightarrow \frac{{48}}{{x + y}} + \frac{{48}}{{x - y}} = 14\,.\,.\,.\,.\,.\,\left( {\text{i}} \right) \cr & {\text{and, }}\frac{4}{{x + y}} = \frac{3}{{x - y}} \cr & \frac{{x - y}}{{x + y}} = \frac{3}{4} = \frac{6}{8} \cr & {\text{From equation }}\left( {\text{i}} \right), \cr & \frac{{48}}{8} + \frac{{48}}{6} = 14 \cr & 14 = 14\,\,\,\left( {{\text{Satisfied}}} \right) \cr & {\text{So, }}x - y = 6 \cr & \,\,\,\,\,\,\,\,\,x + y = 8 \cr & \,\,\,\,\,\overline {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = 1\,\,} {\text{km/hr}} \cr} $$

$$\eqalign{ & {\text{B}} + {\text{W}} = 12{\text{ km/h}} \cr & {\text{B}} - {\text{W}} = 10{\text{ km/h}} \cr & \overline {\,{\text{B}} = \frac{{22}}{2} = 11{\text{ km/h}}\,} \cr} $$

Let speed of boat is x and stream is y km/hr According to the question $$\eqalign{ & \frac{{25}}{{x - y}} + \frac{{39}}{{x + y}} = 8 \cr & \frac{{35}}{{x - y}} + \frac{{52}}{{x + y}} = 11 \cr} $$ By hit and trail method x = 9, y = 4 it satisfy the both equations Speed of stream = 4 km/hr