Question

A motorboat whose speed is $24km/hr$ in still water takes $1hour$ more to go $32km$ upstream than to return downstream to the same spot. Find the speed of the stream.

Answer 1

Step 1: Finding the speed of motorboat moving in downstream and upstream

Given that:

Total Distance$=32km$
Speed in Still Water$=24km/h$

Let the speed of the stream be $‘x’km/h$
then, Speed of moving motorboat in upstream$=24-x$
Speed of moving motorboat in downstream $=24+x$

Step 2: Finding the time taken for the upstream and downstream journey

We know that,

$Timetaken=\frac{dis\tan ce}{speed}$

So, for the upstream journey

$Timetaken=\frac{32}{24-x}hours$

For downstream journey

$Timetaken=\frac{32}{24+x}hours$

Step 3: Forming the equation to find the speed of stream

Given the difference between timings $=1hr$

So,
Time of upstream journey $=$Time of downstream journey $+1hr$

Hence the equation becomes

$$\begin{gathered} \left(\frac{32}{24-x}\right)-\left(\frac{32}{24+x}\right)=1 \\ \Rightarrow 32\left[\left(\frac{1}{24-x}\right)-\left(\frac{1}{24+x}\right)\right]=1 \\ \Rightarrow \left(\frac{1}{24-x}\right)-\left(\frac{1}{24+x}\right)=\frac{1}{32} \\ \Rightarrow \frac{\left(24+x\right)-\left(24-x\right)}{\left(24-x\right)\left(24+x\right)}=\frac{1}{32} \\ \Rightarrow \frac{24+x-24+x}{576+24x-24x-x^2}=\frac{1}{32} \\ \Rightarrow \frac{2x}{576-x^2}=\frac{1}{32} \\ \Rightarrow 64x=576-x^2 \\ \Rightarrow x^2+64x-576=0 \end{gathered}$$

Step 4: Solving the equation to find $x$ i.e. the speed of the stream

On factorizing we get

$$\begin{gathered} x^2+64x-576=0 \\ \Rightarrow x^2+72x-8x-576=0 \\ \Rightarrow x\left(x+72\right)-8\left(x+72\right)=0 \\ \Rightarrow \left(x-8\right)\left(x+72\right)=0 \\ \Rightarrow x=8orx=-72 \end{gathered}$$

Speed cannot be negative

Hence, $x=8$

Therefore, the speed of the stream is $8km/hr$.