Question

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

Answer 1

Hint: First of all, let the speed of stream be x. Now, the speed of the boat travelling upstream will be $18 - x$ and the speed of the boat travelling downstream will be $18 + x$. Now, as the time taken by boat travelling upstream is 1 hour more than travelling downstream, form an equation and use the formula $Time = \dfrac{\text{distance}}{\text{speed}}$.

Complete step-by-step solution:
In this question, we are given that the speed of the boat In still water is 18 km/hr and the time taken by the boat to travel upstream is 1 hour more than travelling downstream. The upstream distance is 24kms. Now, we need to find out the speed of the stream.
Now, let the speed of the stream be x.
So, while travelling upstream, the speed of the boat will reduce due to the speed of the stream. Therefore,
$ \Rightarrow $ Speed of boat travelling upstream $ = $ Speed of boat in still water – Speed of stream
$ \Rightarrow $ Speed of boat travelling upstream $ = 18 - x$
Now, while travelling downstream, the speed of the boat will get a boost due to the speed of the stream flowing downwards too. Therefore,
$ \Rightarrow $ Speed of boat travelling downstream $ = $ Speed of boat in still water + Speed of stream
$ \Rightarrow $ Speed of boat travelling downstream $=18+x$
Now, the boat travels a distance of 24kms upstream, so it will travel a distance of 24kms only while travelling downstream. Therefore,
$ \Rightarrow $ Upstream Distance $ = 24km$
$ \Rightarrow $ Downstream Distance $ = 24km$
Now, it is given that the time taken by boat to travel 24kms upstream is 1 hr more than the time taken by the boat to travel downstream. Therefore,
$ \Rightarrow $ Time taken to travel upstream $ = $ Time taken to travel downstream +1
Now, we know that
$ \Rightarrow Speed = \dfrac{\text{distance}}{\text{time}} \\
   \Rightarrow Time = \dfrac{\text{distance}}{\text{speed}} $
Therefore, using this formula, we get
$\Rightarrow \dfrac{\text{Upstream - distance}}{\text{Upstream - speed}} = \dfrac{\text{Downstream - distance}}{\text{Downstream - speed}} + 1 \\
   \Rightarrow \dfrac{{24}}{{18 - x}} = \dfrac{{24}}{{18 + x}} + 1 $
Taking LCM on the RHS, we get
$ \Rightarrow \dfrac{{24}}{{18 - x}} = \dfrac{{24 + 18 + x}}{{18 + x}} \\
   \Rightarrow \dfrac{{24}}{{18 - x}} = \dfrac{{42 + x}}{{18 + x}}$
Now, cross multiplying, we get
$ \Rightarrow 24\left( {18 + x} \right) = \left( {42 + x} \right)\left( {18 - x} \right) \\
   \Rightarrow 432 + 24x = 756 - 42x + 18x - {x^2} $
Gathering all the like terms, we get
$\Rightarrow {x^2} + 24x + 42x - 18x + 432 - 756 = 0 \\
   \Rightarrow {x^2} + 48x - 324 = 0 $
Now, we are going to solve this equation using the quadratic formula. The quadratic formula is
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here, a = 1, b = 48 and c = -324.
\[ \Rightarrow x = \dfrac{{ - 48 \pm \sqrt {{{48}^2} - 4\left( 1 \right)\left( { - 324} \right)} }}{{2\left( 1 \right)}} \\
   \Rightarrow x = \dfrac{{ - 48 \pm \sqrt {2304 + 1296} }}{2} \\
   \Rightarrow x = \dfrac{{ - 48 \pm \sqrt {3600} }}{2} \\
   \Rightarrow x = \dfrac{{ - 48 \pm 60}}{2} \]
\[ \Rightarrow x = \dfrac{{ - 48 + 60}}{2} \\
   \Rightarrow x = \dfrac{{12}}{2} \\
   \Rightarrow x = 6km/hr \]
OR
\[\Rightarrow x = \dfrac{{ - 48 - 60}}{2} \\
   \Rightarrow x = \dfrac{{ - 108}}{2} \\
   \Rightarrow x = - 54 \]
But, speed cannot be negative, so the speed of the stream will be \[x = 6km/hr\].

Note: Note that here the most important is deciding the speeds of the boat while travelling upstream and downstream. When the boat travels upstream, the water is going to oppose the boat so the speed is always going to be less than that of still water. And when the boat is travelling downstream, the speed of the boat is always going to be more than the actual speed as it gets help from the stream that moves downward too.