Question

A crew can row a certain course up stream in $84$ minutes; they can row the same course down stream in $9$ minutes less than they could row it in still water : how long would they take to row down with the stream?

Answer 1

Let the speed of the boat in still water be $x$

And speed of the stream be $y$

Also, let the length of the course is $d$

The speed of the boat upstream is $x-y$

Time taken for the boat to row upstream is $\frac{d}{x-y}=84\left(1\right)$

Time taken in still water is $\frac{d}{x}$

The speed of the boat in downstream is $x+y$

Time taken in downstream is $\frac{d}{x+y}$

According to the question, $\frac{d}{x}-\frac{d}{x+y}=9$

$\Rightarrow \frac{dy}{x\times (x+y)}=9\left(2\right)$

Dividing $\left(2\right)$ from $\left(1\right)$ we get,

$\left(\frac{y}{x}\right)\times \frac{x-y}{x+y}=\frac{3}{28}$

$\Rightarrow \left(\frac{y}{x}\right)\times \left(\frac{1-\left(\frac{y}{x}\right)}{1+\left(\frac{y}{x}\right)}\right)=\frac{3}{28}$

Let $\left(\frac{y}{x}\right)=k$

$\Rightarrow k\times \frac{1-k}{1+k}=\frac{3}{28}$

$\Rightarrow 28k^2-25k+3=0$

$\Rightarrow k=\frac{1}{7}\text{or}k=\frac{3}{4}$

From equation $\left(1\right),\frac{d}{x-y}=84$

$\Rightarrow \frac{d}{x}\times \frac{1}{1-\left(\frac{y}{x}\right)}=84$

$\Rightarrow \frac{d}{x}\times \frac{1}{1-k}=84$

Putting $k=\frac{1}{7},\frac{d}{x}=84\times \frac{6}{7}=72\text{minutes}$

So, time taken to row in downstream $=\frac{d}{x+y}=\frac{d}{x}\times \frac{1}{1+k}=72\times \frac{7}{8}=63\text{minutes}$

Putting $k=\frac{3}{4},\frac{d}{x}=84\times \frac{1}{4}=21\text{minutes}$

So, time taken to row in downstream $=\frac{d}{x+y}=\frac{d}{x}\times \frac{1}{1+k}=21\times \frac{4}{7}=12\text{minutes}$