Question

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.

Answer 1

Hint:
1) Upstream means 'opposite to / against the direction of the flow' and downstream means 'in the same / along the direction of the flow'.
2) Upstream speed is less than the speed in the still water and the downstream speed is more than the speed in still water.
3) If the speed of the boat in still water is x km/h and the speed of the stream is w km/h, then the upstream speed is (x - w) km/h and the downstream speed is (x + w) km/h.
4) $\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}$ .

Complete step by step solution:
Let's say that the speed of the motorboat is x = 18 km/h, the speed of the stream is w km/h and the distance is d = 24 km.
We know that $\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}$ .
∴ Upstream speed is $x-w=18-w\text{ km/h}$ .
Upstream time = $\dfrac{\text{Distance}}{\text{Speed}}=\dfrac{24}{18-w}\text{ hours}$ .
Downstream speed is $x+w=18+w\text{ km/h}$ .
Downstream time = $\dfrac{\text{Distance}}{\text{Speed}}=\dfrac{24}{18+w}\text{ hours}$ .
According to the question:
 $\dfrac{24}{18+w}+1=\dfrac{24}{18-w}$
On equating the denominators, we get:
⇒$\dfrac{24(18-w)}{(18+w)(18-w)}+\dfrac{(18+w)(18-w)}{(18+w)(18-w)}=\dfrac{24(18+w)}{(18-w)(18+w)}$
On cancelling the denominators [multiplying by $(18+w)(18-w)$ ], we get:
⇒ $24(18-w)+(18+w)(18-w)=24(18+w)$
On multiplying the terms, we get:
⇒ $432-24w+324-{{w}^{2}}=432+24w$
⇒ ${{w}^{2}}+48w-324=0$
On splitting the middle term, we get:
⇒ ${{w}^{2}}+54w-6w-324=0$
On separating the common factors, we get:
⇒ $w(w+54)-6(w+54)=0$
⇒ $(w+54)(w-6)=0$
Since the product of two terms is 0, one of them must be definitely 0:
⇒ $w+54=0$ OR $w-6=0$
⇒ $w=-54$ OR $w=6$

Since the speed of the stream is not negative, we have speed of the stream = $w=6\text{ km/h}$ .

Note:
1) For a constant speed, the distance and the time are directly proportional.
2) The similar concept can be applied to other assisted motions such as on an escalator.
3) If the speed of the boat in still water is x km/h and the speed of the stream is w km/h:
3.1) For an equal distance up and down the stream, the average speed throughout the journey is: $\dfrac{(\text{Speed in Upstream})\times (\text{Speed in Downstream})}{\text{Speed in Still Water}}=\dfrac{({{x}^{2}}-{{w}^{2}})}{x}\text{ km/h}$ .
3.2) If it takes t hours more to go upstream than to go downstream for the same distance, then the distance is calculated by: $\dfrac{t({{x}^{2}}-{{w}^{2}})}{2w}\text{ km}$ .
3.3) If it takes total t hours to row to a place and come back (upstream and downstream both together), then the distance between the two places is: $\dfrac{t({{x}^{2}}-{{w}^{2}})}{2x}\text{ km}$ .
3.4) A boat rows a certain distance downstream in ${t}_{1}\text{ hours}$ and returns the same distance upstream in ${t}_{2}\text{ hours}$ .
3.5) The speed of the boat in still water will be calculated by: $\dfrac{w({{t}_{2}}+{{t}_{1}})}{{{t}_{2}}-{{t}_{1}}}\text{ km/h}$ .
3.6) The speed of the stream will be calculated by: $\dfrac{x({{t}_{2}}-{{t}_{1}})}{{{t}_{2}}+{{t}_{1}}}\text{ km/h}$ .
3.7) A boat takes n times as long to row upstream as to row downstream the river. Then: $x=w\left( \dfrac{n+1}{n-1} \right)$.