Question

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

Answer 1

Hint: First assume the speed of the stream. Then, find the speed of the boat upstream and downstream. While going upstream, the flow of the stream is opposite to the motion of the boat and while going downstream it’s along the direction of the boat. Thus, use the formula for finding the time $t = \dfrac{d}{s}$. Substitute the value and solve the quadratic equation obtained to get the speed of the stream.

Complete step-by-step solution:
In the question we are told about a motorboat whose speed is 18 km/hr, in still water, it takes 1 hour more to go 24 km than to return downstream to the same spot. Now we have to find the speed of the stream.
As we are given, the speed of a boat in still water is equal to 18 km/hr. Let's suppose the speed of the stream is s km/hr.
So, as we know,
Speed of upstream = Speed of a boat in still water – Speed of stream
which can be written as,
$ \Rightarrow $ Speed of upstream \[ = 18-s\]
Now for downstream,
Speed of downstream = Speed of a boat in still water + Speed of steam
which can be written as,
$ \Rightarrow $ Speed of downstream $ = 18 + s$
Now, according to the question we are told,
$ \Rightarrow $ Time taken for upstream = Time taken to cover downstream + 1
The time taken by boat upstream will be,
$ \Rightarrow $ Time taken for upstream $ = \dfrac{{{\rm{Distance\ of\ upstream}}}}{{{\rm{Speed\ of\ upstream}}}}$
The time taken by boat downstream will be,
$ \Rightarrow $ Time taken for downstream $ = \dfrac{{{\rm{Distance\ of\ downstream}}}}{{{\rm{Speed\ of\ downstream}}}}$
We know that distance of upstream and downstream is equal to 24 km so we can write the equation as,
$ \Rightarrow \dfrac{{24}}{{18 + s}} = \dfrac{{24}}{{18 - s}} + 1$
Move variable part on one side,
$ \Rightarrow \dfrac{{24}}{{18 + s}} - \dfrac{{24}}{{18 - s}} = 1$
Now taking LCM we get,
$ \Rightarrow \dfrac{{24\left( {18 + s - 18 + s} \right)}}{{\left( {18 + s} \right)\left( {18 - s} \right)}} = 1$
Simplify the terms,
$ \Rightarrow \dfrac{{24 \times 2s}}{{\left( {18 + s} \right)\left( {18 - s} \right)}} = 1$
On cross multiplication we get,
$ \Rightarrow 48s = 324 - {s^2}$
Move all terms on one side,
$ \Rightarrow {s^2} + 48s - 324 = 0$
Now using middle term factor, we get,
$ \Rightarrow {s^2} - 6s + 54s - 324 = 0$
Take common from the terms,
$ \Rightarrow s\left( {s - 6} \right) + 54\left( {s - 6} \right) = 0$
Again, taking common terms, we get
$ \Rightarrow \left( {s + 54} \right)\left( {s - 6} \right) = 0$
So, the value of s will be,
$ \Rightarrow s = - 54,6$
Since the speed of the stream cannot be negative.
$\therefore s = 6$

Hence, the speed of the stream is 6 km/hr.

Note: If we have the speed of the boat in still water as $u$ km/h and the speed of the stream is $v$ km/h, some students often make this mistake of taking upstream speed as $\left( {u - v} \right)$ km/h. The correct upstream speed of the boat is $\left( {v - u} \right)$ km/h. So, this mistake must be avoided. Students can remember it like, while going upstream, the stream opposes the flow of the boat so the speed of the stream must get subtracted from the speed of the boat and not the other way around.