**Problem**

A man can row 77 km and back in 14 hours. If he can row 6.5 km with the stream at the same time as 4.8 km against the stream, find the rate of the stream.

**Solution**

x = rate of boatman in still water

y = rate of stream

A man can row 77 km and back in 14 hours

$t_{\text{going downstream}} + t_{\text{going upstream}} = 14$

$\dfrac{77}{x + y} + \dfrac{77}{x - y} = 14$

$\dfrac{11}{x + y} + \dfrac{11}{x - y} = 2$ ← Equation (1)

He can row 6.5 km with the stream at the same time as 4.8 km against the stream

$t_{\text{6.5 km with the stream}} = t_{\text{4.8 km against the stream}}$

$\dfrac{6.5}{x + y} = \dfrac{4.8}{x - y}$

$6.5(x - y) = 4.8(x + y)$

$6.5x - 6.5y = 4.8x + 4.8y$

$1.7x = 11.3y$

$x = \frac{113}{17}y$

Substitute x = 113y/17 to Equation (1)

$\dfrac{11}{\frac{113}{17}y + y} + \dfrac{11}{\frac{113}{17}y - y} = 2$

$\dfrac{11}{\frac{130}{17}y} + \dfrac{11}{\frac{96}{17}y} = 2$

$\dfrac{187}{130y} + \dfrac{187}{96y} = 2$

$\dfrac{187}{130} + \dfrac{187}{96} = 2y$

$y = 1.693 ~ \text{kph}$ *answer*