Samantha swam upstream for some distance in one hour. She then swam downstream the same river for the same distance in only 12 minutes. If the river flows at 4 mph, how fast can Samantha swim in still water?

recall your d = rt, distance = rate * time.s = Samantha's speed rateone thing to bear in mind is that, when Samantha is going upstream, she's not really going "s" mph fast, because she's going against the current, and the current's rate is 4 mph and subtracting speed from her, then she's really going "s - 4" fast.Likewise, when she's going downstream because she's going with the current, the current is adding speed to her, so she's going "s + 4" fast.Now, let's say the distance she covered both ways was the same "d" miles.Let's recall that since there are 60 minutes in 1 hour, 12 minutes is 12/60 = 1/5 of an hour.[tex]\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Upstream&d&s-4&1\\ Downstream&d&s+4&\frac{1}{5} \end{array}\qquad \implies \begin{cases} d=(s-4)(1)\\ d=(s+4)\left( \frac{1}{5} \right) \end{cases} \\\\[-0.35em] ~\dotfill[/tex][tex]\bf \stackrel{\textit{from 1st equation}}{\boxed{d}=s-4}\qquad \qquad \stackrel{\textit{substituting on the 2nd equation}~\hfill }{\boxed{s-4}=\cfrac{s+4}{5}\implies 5s-20=s+4} \\\\\\ 4s-20=4\implies 4s=24\implies s=\cfrac{24}{4}\implies s=6[/tex]