A motorboat that travels with a speed of 20 km/hour in still water has traveled 20 km against the current and 180 km with the current, having spent 8 hours on the entire trip. Find the speed of the current of the river.

recall your d = rt.  distance = rate * time.

let's say the speed of the current is "c".

keeping in mind that, when the boat is going against the current, the boat is not really going at 2kmh, but instead at " 20 - c ", because the current is subtracting speed from it.

Likewise when is going with the current, is going at a speed of " 20 + c ".

if the boat took say "t" hours on the way over, on the way back it took the slack from the whole 8 hours and "t", namely it took " 8 - t " hours.

[tex]\bf \begin{array}{lccclll} &\stackrel{km}{distance}&\stackrel{kmh}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ \textit{against the current}&20&20-c&t\\ \textit{with the current}&180&20+c&8-t \end{array} \\\\\\ \begin{cases} 20=(20-c)(t)\implies \frac{20}{20-c}=t\\ 180=(20+c)(8-\stackrel{\downarrow }{t})\\ --------------\\ 180=(20+c)\left( 8- \stackrel{\downarrow }{\frac{20}{20-c}}\right) \end{cases} \\\\\\ 180=(20+c)\left( \cfrac{(160-8c)-20}{20-c} \right)[/tex]

[tex]\bf 180=(20+c)\left( \cfrac{140-8c}{20-c} \right)\implies 180(20-c)=(20+c)(140-8c) \\\\\\ 3600c-180c=2800-160c+140c-8c^2 \\\\\\ 3600c-180c=2800-20c-8c^2 \\\\\\ 3600c-180c-2800+20c+8c^2=0\implies 8c^2-160c+800=0 \\\\\\ c^2-20c+100=0\implies (c-10)(c-10)=0\implies \boxed{c=10}[/tex]