A man can row upstream a distance of \( \Large \frac{2}{3} \) km in 10 minutes and returns the same distance downstream in 5 minutes. Ratio of man's speed in still water and that of the stream will be

\( \Large v_{1} \) - \( \Large v_{2} \) = \( \Large \frac{\frac{2}{3}km}{10 min} \)

Therefore, \( \Large v_{1} \) - \( \Large v_{2} \) = \( \Large \frac{2}{30} \)km/min ...(i)

and \( \Large v_{1} \) + \( \Large v_{2} \) = \( \Large \frac{\frac{2}{3}km}{5 min} \)

= \( \Large \frac{2}{15} \)km/min ...(ii)

Solving equation (i) and (ii), we get

\( \Large V_{1} = \frac{\frac{2}{30}+\frac{2}{15}}{2} \) =\( \Large \frac{2+4}{60} = \frac{6}{60} = \frac{1 km}{10 min} \)

and \( \Large V_{2} = \frac{\frac{2}{15}-\frac{2}{30}}{2} \) = \( \Large \frac{4-2}{60} = \frac{2}{60} = \frac{1 km}{30 min} \) 

Therefore, \( \Large \frac{V_{1}}{V_{2}} = \frac{1}{10} \times \frac{30}{1} \)

= 3 : 1