Question

A boat goes 12 km upstream and 40 km downstream in 8hrs. It can go 16 km upstream and 32km downstream at the same time. Find the speed of the boat in still water and the speed of the stream.
A. Speed of boat in still water = 9km/hr and speed of stream 6km/hr.
B. Speed of boat in still water = 6km/hr and speed of stream 2km/hr.
C. Speed of boat in still water = 14km/hr and speed of stream 32km/hr.
D. Speed of boat in still water = 7km/hr and speed of stream 23km/hr.

Answer 1

Hint: We will be using the concepts of speed, distance and time to solve the problem. We will first assume some value of speed of boat in still water and speed of stream then form the equations as per the conditions given in the question to solve the problem further.

Complete step by step answer:
Now, we will first let the speed of the boat in still water be b km/hr. Speed of the stream is s km/hr.
Now, we have been given that the time taken by boat to go 12 km upstream and 40 km downstream is 8 hrs.
Now, we know the speed of boat in stream = speed of boat in still water + speed of stream.
So, the speed of boat in upstream = b – s
The speed of boat in downstream = b + s
Also, we know that $\text{speed}=\dfrac{\text{distance}}{\text{time}}$. Therefore, according to question we have been given that a boat goes 12 km upstream and 40 km downstream in 8hrs so using the distance formula written above and taking the speed of the boat in upstream and downstream as explained above we have ,
$\dfrac{12}{b-s}+\dfrac{40}{b+s}=8...........\left( 1 \right)$
Now, similarly we have been given that the boat can go 16 km upstream and 32 km downstream at the same time. So, we have,
$\dfrac{16}{b-s}+\dfrac{32}{b+s}=8...........\left( 2 \right)$
Now, we will equate equation (1) and (2). So, that we have,
$\dfrac{12}{b-s}+\dfrac{40}{b+s}=\dfrac{16}{b-s}+\dfrac{32}{b+s}$
Now, on rearranging terms we have,
$\begin{align}
  &\Rightarrow \dfrac{40-32}{b+s}=\dfrac{16-12}{b-s} \\
 &\Rightarrow \dfrac{8}{b+s}=\dfrac{4}{b-s} \\
 & \Rightarrow \dfrac{2}{b+s}=\dfrac{1}{b-s} \\
\end{align}$
Now, we will cross – multiply the above equation so we have,
$\begin{align}
  &\Rightarrow 2\left( b-s \right)=b+s \\
 &\Rightarrow 2b-2s=b+s \\
 &\Rightarrow b=3s.........\left( 3 \right) \\
\end{align}$
Now, we will put $b=3s$ in equation (1). So, we have,
$\begin{align}
  & \dfrac{12}{3s-s}+\dfrac{40}{3s+s}=8 \\
 &\Rightarrow \dfrac{12}{2s}+\dfrac{40}{4s}=8 \\
 &\Rightarrow \dfrac{6}{s}+\dfrac{10}{s}=8 \\
\end{align}$
Now, we will take LCM of the above equation as s and further simplify it to find the value of s
$\begin{align}
  & \dfrac{16}{s}=8 \\
 & \Rightarrow 2=s \\
 &\Rightarrow s=2km/hr \\
\end{align}$
On substituting ‘s’ in (3) we have,
$\begin{align}
  & b=2\times 3 \\
 & =6km/hr \\
\end{align}$
So, the speed of boat in still water = 6 km/hr,
And the speed of the stream is 2 km/hr.

So, the correct answer is “Option B”.

Note: To solve these types of questions it is important to know that in upstream the speed of boat will be less than the speed of boat in still water and in downstream the speed is more due to the added speed of stream.