Hello, rossco21!

We will use: .$\displaystyle \text{Time} \:=\:\frac{\text{Distance}}{\text{Speed}} $

Calum was floating down a river on a raft, when a half-mile downstream,

his brother Duncan set off in a canoe. .Duncan paddled downstream as

quickly as he could, then turned round and paddled back again, still at

his best pace. He arrived back at his starting point just as Calum floated by.

Assuming Duncan's best pace in still water is 10 times that of the river current,

how far did he paddle? Code:

C ½
* - - - - *
D x P
+ - - - - * → → → → → → → → *
* ← ← ← ← ← ← ← ← *
x

Let $\displaystyle r$ be the speed of the river's current.

Calum starts at $\displaystyle C$ and drifts a half-mile to $\displaystyle D.$

. . At $\displaystyle r$ mph, this takes him: .$\displaystyle \frac{\frac{1}{2}}{r} \:=\:\frac{1}{2r}$ hours.

Duncan's speed is $\displaystyle 10r$ mph.

Paddling *with* the current, his speed is: $\displaystyle 10r + r \,=\,11r$ mph.

Paddling *against* the current, his speed is: $\displaystyle 10r - r \,=\,9r$ mph.

Duncan paddles $\displaystyle x$ miles downstream from $\displaystyle D$ to $\displaystyle P$ at $\displaystyle 11r$ mph.

. . This takes him: .$\displaystyle \frac{x}{11r}$ hours.

Duncan paddles $\displaystyle x$ miles upstream back to $\displaystyle D$ at $\displaystyle 9r$ mph.

. . This takes him: .$\displaystyle \frac{x}{9r}$ hours.

Duncan's total time is: .$\displaystyle \frac{x}{11r} + \frac{x}{9r} \:=\:\frac{20x}{99r}$ hours.

Since Calum and Duncan meet at point $\displaystyle D$, their times are equal.

Our equation is: .$\displaystyle \frac{20x}{99r} \;=\;\frac{1}{2r}$

Multiply by $\displaystyle 198r\!:\;\;40x \:=\:99\quad\Rightarrow\quad x \:=\:\frac{99}{40}$ miles.

Therefore, Duncan paddled a total of: .$\displaystyle 2x \:=\:2\left(\frac{99}{40}\right) \:=\:4.95$ miles.