A boat covers a certain distance downstream in 1 hour while it comes back in \[1\dfrac{1}{2}\] hours. If the speed of the stream is 3 kilometre per hour. What is the speed of the boat in still water?

Hint: The appropriate approach to the above question is by solving it using the boats and streams method in which we need to assume the value of speed of boat in still water to be some variable and as given the time of upstream and downstream we can get the distance travelled upstream and downstream by using the velocity, distance, time relationship and then equate them.

Complete step-by-step answer:

Formula related to boats and streams:

Let the speed of boat in still water is x km/h and speed of the stream is y km/h then,
\[\begin{align}
  & \text{Speed in downstream }=\left( x+y \right)\text{ km/h}...............\left( 1 \right) \\
 & \text{Speed in upstream }=\left( x-y \right)\text{ km/h}....................\left( 2 \right) \\
\end{align}\]

Speed in downstream > Speed of boat in still water................(3)
Speed in upstream < Speed of boat in still water

Now in the question given that:
The speed of the stream is 3 km/h
Time of downstream is 1 h
Time of upstream is \[1\dfrac{1}{2}\] h

Let us assume that the speed of the boat in still water as x km/h
Time of upstream as tu h
Time of downstream as td h
Speed in downstream as vd km/h
Speed in upstream as vu km/h

As we already know that the relation between velocity, time and distance is expressed as:
\[\begin{align}
  & \text{velocity}=\dfrac{\text{distance }}{\text{time}} \\
 & \text{distance}=\text{velocity}\times \text{time} \\
\end{align}\]

From the equations (1) and (2) listed above we get,
Speed of downstream \[=x+3\]
Speed of upstream \[=x-3\]

Now, as the distance travelled will be equal in upstream and downstream we get,
\[\Rightarrow \left( x+y \right){{t}_{d}}=\left( x-y \right){{t}_{u}}\]

Now, on substituting the corresponding values we get,
\[\begin{align}
  & \Rightarrow \left( x+3 \right)\times 1=\left( x-3 \right)\times 1\dfrac{1}{2} \\
 & \Rightarrow \left( x+3 \right)\times 1=\left( x-3 \right)\times \dfrac{3}{2} \\
\end{align}\]
\[\Rightarrow x+3=\dfrac{3}{2}x-\dfrac{9}{2}\]

Now, on rearranging the terms we get,
\[\begin{align}
  & \Rightarrow 3+\dfrac{9}{2}=\dfrac{3}{2}x-x \\
 & \Rightarrow \dfrac{15}{2}=\dfrac{x}{2} \\
 & \therefore x=15km/h \\
\end{align}\]

Hence, the speed of the boat in still water is 15 kilometre per hour.

Note: It is important to note that the distance travelled upstream is equal to the distance travelled downstream because given in the question relating time with the distance. Which means that the time downstream to reach a certain point is given and the time upstream to reach the same point is given.
While equating the distance it is important to express the distance in terms of speed of upstream, downstream and time of upstream, downstream because we know the values of upstream and downstream time and the value of the speed of the stream . So, we have only one unknown here which is the speed of the boat in still water.