A man row to a place 48 km distant and back on 14 hours text. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. The rate of the stream is?

11 km/hr
21.5 km/hr
31.8 km/hr
43.5 km/hr

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Suppose he moves 4 km downstream in x hours . \\ Then , Speed downstream \\ = (\frac{4}{x}) k m / h r \\ Speed upstream \\ = (\frac{3}{x}) k m / h r \\ ∴ \frac{48}{(\frac{3}{x})} + \frac{48}{(\frac{4}{x})} = 14 o r x = \frac{1}{2} \\ S o, Speed downstream \\ =  7 k m / h r \\ Speed uptream  =  6 k m / h r \\ Rate of the stream \\ = \frac{1}{2} (8 - 6) k m / h r \\ = 1 k m / h r \end{aligned}

$\eqalign{ & {\text{Suppose he moves 4 km downstream in }}x{\text{ hours}}{\text{.}} \cr & {\text{Then , Speed downstream}} \cr & {\text{ = }}\left( {\frac{4}{x}} \right)km/hr \cr & {\text{Speed upstream}} \cr & {\text{ = }}\left( {\frac{3}{x}} \right)km/hr \cr & \therefore \frac{{48}}{{\left( {\frac{3}{x}} \right)}} + \frac{{48}}{{\left( {\frac{4}{x}} \right)}} = 14\,\,\,or\,\,\,x = \frac{1}{2} \cr & So,\,{\text{Speed downstream}} \cr & {\text{ = 7 }}km/hr \cr & {\text{Speed uptream = 6 }}km/hr \cr & {\text{Rate of the stream}} \cr & {\text{ = }}\frac{1}{2}\left( {8 - 6} \right)km/hr \cr & = 1\,km/hr \cr}$