Question

A boat can travel 47 mph in still water. If it travels 280 miles with the current in the same length of time it travels 190 miles against the current, what is the speed of the current?

Answer 1

b = speed of the boat in still water = 47

c = speed of the current

when going Upstream, the boat is not really going "47 mph" fast, is really going slower, is going "47 - c", because the current is subtracting speed from it, likewise, when going Downstream the boat is not going "47 mph" fast, is really going faster, is going "47 + c", because the current is adding its speed to it.

`{\Large \begin{array}{llll} \underset{distance}{d}=\underset{rate}{r} \stackrel{time}{t} \end{array}} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Downstream&280&47+c&h\\ Upstream&190&47-c&h \end{array}\hspace{5em} \begin{cases} 280=(47+c)(h)\\\\ 190=(47-c)(h) \end{cases} \\\\[-0.35em] ~\dotfill`

`\stackrel{\textit{using the 1st equation}}{280=(47+c)h}\implies \cfrac{280}{47+c}=h \\\\\\ \stackrel{\textit{using the 2nd equation}}{190=(47-c)h}\implies \stackrel{\textit{substituting from above}}{190=(47-c)\left( \cfrac{280}{47+c} \right)} \\\\\\ 190(47+c)=(47-c)280\implies 8930+190c=13160-280c \\\\\\ 190c=4230-280c\implies 470c=4230\implies c=\cfrac{4230}{470}\implies \boxed{c=9}`