If the speed of a boat in still water is v km/h and the speed of stream is u km/h, then the speed of boat in the opposite direction of stream (upstream) = (v - u) km/h.

\[\therefore\] Required speed

= (10 - 5) km/h

= 5 km/h

If the speed of a boat in the direction of stream (downstream) is x km/h and in the opposite direction of stream (upstream) is y km/h, then speed of the boat is \[\frac{x+y}{2}\] km/h

\[\therefore\] Required speed

\[=\frac{20+10}{2}\]

\[=\frac{30}{2}\]

\[=15\] km/h

If the speed of a boat in the direction of stream (downstream) is x km/h and in the opposite direction of stream (upstream) is y km/h, then speed of the stream is \[\frac{x-y}{2}\] km/h

\[\therefore\] Required speed

\[=\frac{20-10}{2}\]

\[=\frac{10}{2}\]

\[=5\] km/h

Let the speed of boat is x km/h and speed of stream is y km/h. To travel \[d_1\] km downstream and \[d_2\] km upstream, the time taken is t hours, then

\[t = \frac{d_1}{x+y} +\frac{d_2}{x-y} \]

\[\therefore\] Required time

\[t = \frac{2}{30+10} +\frac{1}{30-10} \]

\[t = \frac{2}{40} +\frac{1}{20} \]

\[t = \frac{2+2}{40}\]

\[t = \frac{4}{40}\]

\[t = \frac{1}{10}\] hour

\[t = \frac{1}{10} \times 60\] minute

\[=6\] minute

Let the speed of boat be x km/h and speed of stream be y km/h. A boat travels equal distance d upstream as well as downstream in t hours, then

\[d=\frac{t\left(x^2-y^2\right)}{2x}\]

\[\therefore\] Required distance

\[\frac{t\left(x^2-y^2\right)}{2x}\]

\[=\frac{2\left(4^2-2^2\right)}{2\times 2}\]

\[=\frac{2\left(16-4\right)}{4}\]

\[=\frac{2\times 12}{4}\]

\[=6\] km

If a boat takes t hours to travel x km upstream and y km downstream, then,

\[\text{Speed of boat} = \frac{\text{Sum of distances}}{2 \times \text{time}}\]

\[\therefore \text{Required speed} = \frac{4+6}{2 \times 2}\]

\[= \frac{10}{4}\]

\[= 2.5 \] km/h

If a boat takes t hours to travel x km upstream and y km downstream, then,

\[\text{Speed of boat} = \frac{\text{Difference of distances}}{2 \times \text{time}}\]

\[\therefore \text{Required speed} = \frac{6-4}{2 \times 2}\]

\[= \frac{2}{4}\]

\[= \frac{1}{2}\]

\[= \frac{1}{2} \times 60 \] minutes

\[= 30 \] minutes

If a boat takes same time to travel \[d_1\] km downstream and \[d_2\] km upstream, then,

\[\frac{\text{Speed of boat}}{\text{Speed of stream}} =\frac{d_1+d_2}{d_1-d_2}\]

\[\therefore \text{Speed of boat} =\frac{d_1+d_2}{d_1-d_2} \times \text{Speed of stream}\]

\[= \frac{3+2}{3-2} \times 4\]

\[= \frac{5}{1} \times 4\]

\[= 20\] km/h

If a boat covers x km distance in \[t_1\] hours along the direction of stream (downstream) and covers the same distance in \[t_2\] hours against the stream i.e. upstream, then

\[\text{Speed of boat} = \frac{x}{2}\left(\frac{1}{1}+\frac{1}{2}\right)\]

\[\therefore \text{Required speed} =\frac{10}{2}\left(\frac{2+1}{2}\right)\]

\[=\frac{10}{2}\left(\frac{3}{2}\right)\]

\[=\frac{30}{4}\]

\[=7.5\] km/h

If a boat covers x km distance in \[t_1\] hours along the direction of stream (downstream) and covers the same distance in \[t_2\] hours against the stream i.e. upstream, then

\[\text{Speed of stream} = \frac{x}{2}\left(\frac{1}{1}-\frac{1}{2}\right)\]

\[\therefore \text{Required speed} =\frac{10}{2}\left(\frac{2-1}{2}\right)\]

\[=\frac{10}{2}\left(\frac{1}{2}\right)\]

\[=\frac{10}{4}\]

\[=2.5\] km/h

If the speed of a boat or swimmer in still water is a km/h and river is flowing with a speed of b km/h, then average speed in going to a certain place and coming back to starting point is given by

\[\frac{(a+b)(a-b)}{a}\] km/h

\[\therefore \text{Required speed} = \frac{(10+8)(10-8)}{10}\]

\[= \frac{(18)(2)}{10}\]

\[= \frac{36}{10}\]

\[= 3.6\] km/h

Speed of the boat

\[=\frac{\text{Rate downstream + Rate upstream}}{2}\]

\[=\frac{10+4}{2}\]

\[=\frac{14}{2}\]

\[=7\] km/h

Speed of the stream

\[=\frac{\text{Rate downstream - Rate upstream}}{2}\]

\[=\frac{10-4}{2}\]

\[=\frac{6}{2}\]

\[=3\] km/h

Rate downstream

\[= \frac{18}{4} = \frac{9}{2}\] km/h

Rate upstream

\[= \frac{18}{12} = \frac{3}{2}\] km/h

Now, Speed of the boat

\[=\frac{\text{Rate downstream + Rate upstream}}{2}\]

\[\frac{\frac{9}{2}+\frac{3}{2}}{2}\]

\[=\frac{9+3}{2\times 2}\]

\[=\frac{12}{4}\]

\[=3\] km/h

Rate downstream

\[= \frac{18}{4} = \frac{9}{2}\] km/h

Rate upstream

\[= \frac{18}{12} = \frac{3}{2}\] km/h

Now, Speed of the stream

\[=\frac{\text{Rate downstream - Rate upstream}}{2}\]

\[=\frac{\frac{9}{2}-\frac{3}{2}}{2}\]

\[=\frac{6}{4}\]

\[=\frac{3}{2}\]

\[=1.5\] km/h

Method - 1

Let the speed of the current be x kmph. According to the question,

\[\frac{6}{6-x} = 3\]

\[\implies 18 – 3 x = 6 \]

\[\implies 3x = 18 – 6\]

\[\implies x = \frac{12}{3}\]

\[= 4\] km/h

Method - 2

Here, Speed of boat = 6 km/hr, \[t_1 = 3x, t_2 = x\]

\[\frac{\text{Speed of Boat}}{\text{Speed of Stream}} = \frac{t_1 + t_2}{t_1 - t_2}\]

\[\frac{6}{\text{Speed of Stream}} = \frac{3x + x}{3x - x}\]

\[\therefore \text{Speed of Current}\] = 3 km/hr

• Rate downstream = (6 + 1.5) km/h = 7.5 km/h
• Rate upstream = (6 – 1.5) km/h = 4.5 km/h

According to the question,

\[\text{Time} = \frac{\text{Distance}}{\text{Speed}}\]

\[\therefore \text{Required time} = \frac{22.5}{7.5} + \frac{22.5}{4.5}\]

\[= 3 + 5 = 8\] hours.

Let the speed of the stream be x kmph

and both the boats meet after t hours

According to the question,

\[(12 + x ) t + (15 – x ) t = 108\]

\[\implies 12 t + 15 t = 108\]

\[\implies 27 t = 108\]

\[\therefore t = \frac{108}{27} = 4\] hours.

Rate downstream

\[= \frac{20}{4}\]

\[= 5\] km/h

Rate upstream

\[= \frac{9}{3}\]

\[= 3\] km/h

\[\therefore \text{Speed of current}\]

\[=\frac{\text{Rate downstream - Rate upstream}}{2}\]

\[\frac{5-3}{2}\]

\[=1\] km/h