Originally Posted by

**Soroban** Hello, MathLearner!

Here's #2 . . . a challenging problem!

Let $\displaystyle b$ = speed of the boat in still water.

Let $\displaystyle c$ = speed of the current.

The boat went downstream for 20 km at a speed of $\displaystyle (b+c)$ km/hr.

. . This took: $\displaystyle \frac{20}{b+c}$ hours.

The boat went upstrream for 20 km at a speed of $\displaystyle (b-c)$ km/hr.

. . This took: $\displaystyle \frac{20}{b-c}$ hours.

The total trip took 7 hours: .$\displaystyle \frac{20}{b+c} + \frac{20}{b-c} \:=\:7 \quad\Rightarrow\quad 7b^2 - 40b - 7c^2 \:=\:0 \;\;{\color{red}(a)}$

Consider when the boat met the log . . .

The boat had gone downstream 20 km at $\displaystyle (b+c)$ km/hr.

. . This took: $\displaystyle \frac{20}{b+c}$ hours.

Then the boat went upstream 8 km at $\displaystyle (b-c)$ km/hr.

. . This took: $\displaystyle \frac{8}{b-c}$ hours.

Total time: .$\displaystyle \frac{20}{b+c} + \frac{8}{b-c}\text{ hours.}\;\;{\color{blue}[1]}$

During this time, the log traveled 12 km at $\displaystyle c$ km/hr.

. . This took: .$\displaystyle \frac{12}{c}\text{ hours.}\;\;{\color{blue}[2]}$

Equate [1] and [2]: .$\displaystyle \frac{20}{b+c} + \frac{8}{b-c} \:=\:\frac{12}{c}\quad\Rightarrow\quad 12b^2 - 28bc \:=\:0$

Then: .$\displaystyle 4b(3b-7c) \:=\:0 \quad\Rightarrow\quad c \:=\:\frac{3}{7}b\;\;{\color{blue}[3]} \quad\hdots\quad \text{since }b \neq 0$

Substitute into (a): .$\displaystyle 7b^2 - 40b - 7\left(\frac{9}{49}b^2\right) \:=\:0 \quad\Rightarrow\quad 40b^2 - 280b \:=\:0$

. . $\displaystyle 40b(b - 7) \:=\:0 \quad\Rightarrow\quad\boxed{b \:=\:7\text{ km/hr}} $

Substitute into [3]: .$\displaystyle c \:=\:\frac{3}{7}(3) \quad\Rightarrow\quad\boxed{c\:=\:3\text{ km/hr}} $

Therefore, the downstream speed is: .$\displaystyle b + c \:=\:7 + 3 \:=\:{\color{blue}\boxed{10\text{ km/hr}}}$