Question

A boat can travel 15 miles against a current in the same time that it can travel 35 miles with the current. The rate of the current is 2mph. Find the rate of the boat in the still water.

Answer 1

recall your d =rt, or distance = rate * time

so... when the boat is going upstream, is going against the current, hmmmm let's say the boat has a speed rate of "r", we know the current has a rate of 2, so... .if the boat is going against the current, its actual speed is not really "r", but "r - 2", since the current is taking away speed from it

now...going downstream, is going with the current, is not really going "r" fast either, is going "r + 2"

let us notice, the time for upstream travel, is really the same time for the downtream travel, let's say hmm it took "t" long

thus
`\bf \begin{array}{lccclll} &distance&rate&time\\ &-----&-----&-----\\ upstream&15&r-2&t\\ downstream&35&r+2&t \end{array}\\\\ -----------------------------\\\\ \begin{cases} 15=(r-2)t\implies \cfrac{15}{r-2}=\boxed{t}\\\\ 35=(r+2)t\\ ----------\\ 35=(r+2)\left( \boxed{(15)/(r-2)} \right) \end{cases}`

solve for "r"