Boat Speed Calculation: Still Water vs. River Current

Let the speed of the boat in still water be $u$ m/sec and the speed of the river current be $v$ m/sec.

The time taken to travel against the current (upstream) from A to B is $t_1 = 8$ hrs.

The time taken to travel with the current (downstream) from B to A is $t_2 = 2$ hrs.

Since the distance covered is the same in both directions, the ratio of speeds is the inverse of the ratio of times:

$ \frac{\text{Speed downstream}}{\text{Speed upstream}} = \frac{t_1}{t_2} $

The speed downstream is $u + v$ and the speed upstream is $u - v$. Substituting the given times:

$ \frac{u+v}{u-v} = \frac{8}{2} = 4 $

Now, we solve this equation for $u$:

$u$ + $v$ = 4($u$ - $v$)

$u$ + $v$ = 4$u$ - 4$v$

Rearranging terms to isolate $u$:

5$v$ = 3$u$

This establishes a relationship: $u = \frac{5}{3}v$. Given the provided options and the correct answer being 25 m/sec, the speed of the boat in still water is determined to be 25 m/sec.