Question

A boat can travel 29mph in still water. If it travels 342 miles with the current in the same length of time it travels 180 miles against the current, what is the speed of the current?

Answer 1

29 = speed of the boat in still water

c = speed of the current

t = time it took each way

when going Upstream, the boat is not really going "29" fast, is really going slower, is going "29 - c", because the current is subtracting speed from it, likewise, when going Downstream the boat is not going "29" fast, is really going faster, is going "29 + 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}&\\ Upstream&180&29-c&t\\ Downstream&342&29+c&t \end{array}\hspace{5em} \begin{cases} 180=(29-c)(t)\\\\ 342=(29+c)(t) \end{cases} \\\\[-0.35em] ~\dotfill`

`\stackrel{\textit{using the 1st equation}}{180=(29-c)t\implies \cfrac{180}{29-c}=t} \\\\\\ \stackrel{\textit{substituting on the 2nd equation from above}}{342=(29+c)\left( \cfrac{180}{29-c} \right)}\implies \cfrac{342}{29+c}=\cfrac{180}{29-c} \\\\\\ 9918-342c=5220+180c\implies 4698-342c=180c\implies 4698=522c \\\\\\ \cfrac{4698}{522}=c\implies \boxed{9=c}`