A boat goes down the river from point A to point B, which is at the distance of $10$ km from A, and then returns to A. If the actual speed of the boat is $3$ km/h, then it takes $2$ h $30$ min less for the boat to go from A to B than from B to A. What should the actual speed of the boat be for the distance from A to B to be covered in two hours?
Let speed of boat in still water $= u$ kmph
and speed of river $= v$ kmph
Then, speed of boat down river (from A to B) $= \left(u + v\right)$ kmph
and speed of boat up river (from B to A) = $\left(u - v\right)$ kmph
Distance AB $= d \left(AB\right) = 10$ km (given)
Case 1:
Speed of boat in still water $= u = 3$ kmph (given)
Time taken by the boat to go from A to B $= T_1 = \dfrac{10}{u + v} = \dfrac{10}{3 + v}$ hr
Time taken by the boat to go from B to A $= T_2 = \dfrac{10}{u - v} = \dfrac{10}{3 - v}$ hr
As per problem, $T_1 = T_2 - 2.5$ hr
i.e. $\;$ $\dfrac{10}{3 + v} = \dfrac{10}{3 - v} - 2.5$
i.e. $\;$ $\dfrac{10}{3 - v} - \dfrac{10}{3 + v} = 2.5$
i.e. $\;$ $\dfrac{30 + 10v - 30 + 10v}{9 - v^2} = 2.5$
i.e. $\;$ $20v = 22.5 - 2.5 v^2$
i.e. $\;$ $5v^2 + 40v - 45 = 0$
i.e. $\;$ $v^2 + 8v - 9 = 0$
i.e. $\;$ $\left(v + 9\right) \left(v - 1\right) = 0$
i.e. $\;$ $v = -9$ $\;$ or $\;$ $v = 1$
Since speed of river cannot be negative, $\implies$ $v = -9$ is not an acceptable solution
$\therefore \;$ Speed of river $= v = 1$ kmph
Case 2:
Given: Time taken by the boat to cover $d \left(AB\right) = 2$ hr
Speed of boat from A to B $= u + v = u + 1 = \dfrac{\text{distance}}{\text{time}} = \dfrac{10}{2} = 5$ kmph
i.e. $\;$ $u = 5 - 1 = 4$ kmph
Therefore, if the speed of boat in still water $= u = 4$ kmph, then the distance AB will be covered in $2$ hours.