Question

A motor boat whose speed is $18km/hr$ in still water takes 1hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

Answer 1

Hint:Take speed of stream as ‘s’. Find the speed of the boat upstream and downstream. While going upstream, the flow of stream ‘s’ is opposite to the motion of boat and while going downstream it’s along the direction of boat. Thus use the speed formula. Substitute the value and solve the quadratic equation obtained to get the value of ‘s’.

Complete step-by-step answer:
It is said that the motor boat takes extra 1hr to go upstream which is 24 km. As the boat is going upstream, it is opposite to the direction of flow of the stream. Thus, the speed of the boat will reduce.
\[\therefore \] Speed of boat upstream = speed of boat in still water – speed of the stream.
I.e. Speed of the boat upstream = 18-s.
Now the boat should come back to the same spot. That is the boat is going downstream, it is in the direction of flow of the stream. Thus, the speed of motor boat increases.
\[\therefore \] Speed of boat downstream = speed of boat in still water + speed of the stream.
I.e. speed of boat downstream = 18+s.
The time taken for upstream is 1hr more than the time taken to cover downstream.
\[\therefore \]Time taken for upstream = time taken to cover downstream +1.
We know the formula for speed.
\[\text{speed}=\dfrac{\text{Distance}}{\text{time}}\text{ }\therefore \text{time = }\dfrac{ \text{Distance }}{\text{speed}}\]
Here the distance converted is the same i.e. 24 km.
\[\therefore \dfrac{\text{Distance upstream}}{ \text{ speed of boat upstream }}=\dfrac{\text{Distance downstream}}{\text{ speed of boat downstream }}+1.\]
Apply the expression we found and simplify it.
\[\begin{align}
  & \therefore \dfrac{24}{18-s\text{ }}=\dfrac{24}{\text{18+s }}+1 \\
 & \dfrac{24}{18-s\text{ }}=\dfrac{24+18+s}{\text{18+s }} \\
\end{align}\]
By Cross multiplication, we get
\[\begin{align}
  & 24\left( 18+s \right)=\left( 42+s \right)\left( 18-s \right) \\
 & 432+24s=756-42s+18s-{{s}^{2}} \\
 & 432+24s=756-24s-{{s}^{2}} \\
 & {{s}^{2}}+24s+24s+432-756=0 \\
 & {{s}^{2}}+485-324=0 \\
\end{align}\]
This is in the form of a quadratic equation. Let us take 2 terms as a and b.
\[\text {sum of zeroes}=\dfrac{-\text{ coefficient of }x}{\text{coefficient of }{{x}^{2}}}=-\left( \dfrac{-48}{1} \right)=+48=a+b\]
 \[\text{ Product of zeroes}=\dfrac{\text{constant term}}{ \text{ coefficient of }{{x}^{2}}}=\dfrac{-324}{1}=ab\]
\[\begin{align}
  & \therefore a+b=+48 \\
& ab=-324 \\
\end{align}\]
Let us take \[a=-6\] and \[b=+54\]
\[\begin{align}
  & \therefore a+b=-6+54=+48 \\
 & ab=\left( -6 \right)\times \left( +54 \right)=-324 \\
\end{align}\]
Thus we can split the middle term as \[\left( -6+54 \right)=48\].
\[\begin{align}
  & {{s}^{2}}-6s+54s-324=0 \\
 & s\left( s-6 \right)+54\left( s-6 \right)=0 \\
 & \left( s+54 \right)\left( s-6 \right)=0 \\
\end{align}\]
Thus take \[s-6=0\] i.e. $s=6km/hr$.
\[\therefore \] The speed of the stream is $6km/hr$.

Note: Neglected the term \[\left( s+54 \right)=0\]. As we get \[s=-54\]. We assumed ‘s’ as the speed of the boat and speed can never be negative. Remember while forming the equation for upstream and downstream that, the flow of the stream is opposite to the motion of the boat in upstream and vice versa.