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A motorboat whose speed in still water is 18 km/h, takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

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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 stream to be some variable and as given the relation between time of upstream and downstream we can get the relation of velocity of upstream and downstream from that.

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
\[\begin{align}
  & ~\text{Speed of boat in still water }=18\text{ km/h} \\
 & \text{Time of upstream }=1+\text{Time of downstream} \\
\end{align}\]
Let us assume speed of the steam as y km/h
Time of upstream as ${{\text{t}}_{u}}$ h
Time of downstream as ${{\text{t}}_{d}}$ h
Speed in downstream as ${{\text{v}}_{d}}$ km/h
Speed in upstream as ${{\text{v}}_{u}}$ km/h
\[\begin{align}
  & \text{Given }x=18\text{ km/h} \\
 & {{\text{t}}_{u}}=1+{{\text{t}}_{d}} \\
\end{align}\]
\[d=24\text{ km}\]
\[{{\text{t}}_{u}}=\dfrac{24}{{{v}_{u}}}\text{ , }{{\text{t}}_{d}}=\dfrac{24}{{{v}_{d}}}\text{ }\left[ \because \text{velocity}=\dfrac{\text{distance}}{\text{time}} \right]\]
From the above formula related to boats and streams (1) and (2) we can write as
\[\begin{align}
  & {{v}_{d}}=x+y.............\left( 4 \right) \\
 & {{v}_{u}}=x-y..............(5) \\
\end{align}\]
\[\begin{align}
  & \Rightarrow \dfrac{24}{{{v}_{u}}}=1+\dfrac{24}{{{v}_{d}}}\text{ } \\
 & \Rightarrow \dfrac{24}{{{v}_{u}}}-\dfrac{24}{{{v}_{d}}}=1 \\
 & \Rightarrow 24\left( \dfrac{1}{{{v}_{u}}}-\dfrac{1}{{{v}_{d}}} \right)=1 \\
 & \Rightarrow 24\left( \dfrac{{{v}_{d}}-{{v}_{u}}}{{{v}_{u}}\cdot {{v}_{d}}} \right)=1 \\
 & \\
\end{align}\]
Form the above equations (4) and (5) we get
\[\begin{align}
  & {{v}_{d}}-{{v}_{u}}=2y \\
 & {{v}_{u}}\cdot {{v}_{d}}={{x}^{2}}-{{y}^{2}} \\
\end{align}\]
By substituting these in the above step we get
\[\begin{align}
  & \Rightarrow 24\left( \dfrac{2y}{{{x}^{2}}-{{y}^{2}}} \right)=1 \\
 & \Rightarrow 24\left( 2y \right)={{x}^{2}}-{{y}^{2}} \\
 & \Rightarrow 48y={{\left( 18 \right)}^{2}}-{{y}^{2}}\text{ }\left[ \because x=18\text{km/h} \right] \\
 & \Rightarrow {{y}^{2}}+48y-324=0..............\left( 6 \right) \\
\end{align}\]
As we know that solution of a quadratic equation from direct formula
\[a{{x}^{2}}+bx+c=0\]has two roots.........................(7)
\[\alpha =\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\text{ , }\beta \text{=}\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\].................(8)
Let us assume that the two roots of equation (6) as ${{\text{}}_{1}}$ and ${{\text{y}}_{2}}$ now on comparing equation (6) and (7) we get,
\[a=1,b=48,c=-324\]
By substituting these values in equation (8) we get,
\[\begin{align}
  & {{y}_{1}}=\dfrac{-48+\sqrt{{{\left( 48 \right)}^{2}}-4\left( 1 \right)\left( -324 \right)}}{2\times 1} \\
 & \Rightarrow {{y}_{1}}=\dfrac{-48+\sqrt{{{\left( 48 \right)}^{2}}+\left( 4\times 324 \right)}}{2} \\
 & \Rightarrow {{y}_{1}}=\dfrac{-48+\sqrt{2304+1296}}{2} \\
 & \Rightarrow {{y}_{1}}=\dfrac{-48+\sqrt{3600}}{2} \\
 & \Rightarrow {{y}_{1}}=\dfrac{-48+60}{2} \\
 & \Rightarrow {{y}_{1}}=\dfrac{12}{2} \\
 & \therefore {{y}_{1}}=6 \\
\end{align}\]
\[\begin{align}
  & {{y}_{2}}=\dfrac{-48-\sqrt{{{\left( 48 \right)}^{2}}-4\left( 1 \right)\left( -324 \right)}}{2\times 1} \\
 & \Rightarrow {{y}_{2}}=\dfrac{-48-\sqrt{{{\left( 48 \right)}^{2}}+\left( 4\times 324 \right)}}{2} \\
 & \Rightarrow {{y}_{2}}=\dfrac{-48-\sqrt{2304+1296}}{2} \\
 & \Rightarrow {{y}_{2}}=\dfrac{-48-\sqrt{3600}}{2} \\
 & \Rightarrow {{y}_{2}}=\dfrac{-48-60}{2} \\
 & \Rightarrow {{y}_{2}}=\dfrac{-108}{2} \\
 & \therefore {{y}_{2}}=-54 \\
\end{align}\]
Now from equation (3) we know that Speed in downstream > Speed of boat in still water
\[\begin{align}
  & \Rightarrow x+y>x \\
 & \Rightarrow y>x-x \\
 & \therefore y>0 \\
\end{align}\]
\[\therefore y=6\]
So here we only consider the value of ${{\text{y}}_{1}}$ but not ${{\text{y}}_{2}}$
Hence, the speed of the stream is 6 km/h.
Note: Expressing time in terms of velocity and then getting the relation between speed of the upstream, downstream, boat in still water and speed of the stream is one of the crucial parts.
While finding the roots of the quadratic equation obtained in terms of y we can also use factorisation method instead of direct formula method but it takes little time to check which values satisfy the given equation. But both the methods give the same result.
The relation that we considered between speed in downstream and speed of boat in still water i.e., speed in downstream > speed of boat in still water is the most important step in the solution because without considering that inequality we get a different result which is inappropriate.