Solution:
$$\eqalign{
& {\text{Speed of boat}} = x \cr
& {\text{Speed of current}} = y \cr
& {\text{According to the question,}} \cr
& \frac{{40}}{{x + y}} + \frac{{25}}{{x - y}} = \frac{{15}}{2} \cr
& {\text{Multiplying both sides by}}\frac{6}{5}{\text{ and}} \cr
& \frac{{40}}{{x + y}} + \frac{{30}}{{x - y}} = \frac{{18}}{2} \cr
& \frac{{40}}{{x + y}} + \frac{{30}}{{x - y}} = 9\,.\,.\,.\,.\,.\,.\,.\,\left( 1 \right) \cr
& \frac{{48}}{{x + y}} + \frac{{36}}{{x - y}} = 10\,.\,.\,.\,.\,.\,.\,.\,\left( 2 \right) \cr
& \underline {\,\, - \,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{ - 6}}{{x - y}} = - 1 \cr
& x - y = 6\,.\,.\,.\,.\,.\,.\,.\,\left( 3 \right) \cr
& {\text{From equation }}\left( 2 \right), \cr
& \frac{{48}}{{x + y}} + \frac{{36}}{6} = 10 \cr
& \frac{{48}}{{x + y}} = 4 \cr
& x + y = 12\,.\,.\,.\,.\,.\,.\,.\,\left( 4 \right) \cr
& {\text{From equation }}\left( 3 \right){\text{ and }}\left( 4 \right) \cr
& x - y = 6 \cr
& x + y = 12 \cr
& \overline {\,x = 9,\,y = 3\,} \cr
& {\text{Speed of boat}} = 9 \cr} $$
