Question

A boat traveled downstream a distance of 90 miles and then came right back. If the speed of the current was 8mph and the total trip took 6 hrs, find the average speed of the boat relative to the water

Answer 1

Recall your d = rt, distance = rate * time.

keeping in mind that, the distance upstream as well as downstream is the same, 90 miles, let's say the boat has a speed rate of "b", thus when the boat was going upstream, it really wasn't going "b" fast, it was going "b - 8", because the 8mph speed of the current is subtracting from it.  Likewise, when the boat was going downstream, it wasn't going "b" fast either, because the current was adding to it, since it was going with the current, then it was really going "b + 8" fast.

now, let's notice the trip took a total of 6 hours, thus, if the trip downstream took say "t" hours, then the trip upstream took the slack from 6 and "t", that is, it took "6 - t" hours.

$\bf \begin{array}{lccclll} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ Downstream&90&b+8&t\\ Upstream&90&b-8&6-t \end{array} \\\\\\ \begin{cases} 90=t(b+8)\implies \cfrac{90}{b+8}=\boxed{t}\\\\ 90=(6-t)(b-8)\\ ----------\\ 90=\left( 6-\boxed{\cfrac{90}{b+8}} \right)(b-8) \end{cases}$

$\bf 90=\left( \cfrac{6b+48-90}{b+8} \right)(b-8)\implies 90=\left( \cfrac{6b-42}{b+8} \right)(b-8) \\\\\\ 90=\cfrac{(6b-42)(b-8)}{b+8}\implies 90(b+8)=(6b-42)(b-8) \\\\\\ 90b+720=6b^2-48b-42b+336\implies 0=6b^2-180b-384 \\\\\\ 0=b^2-30b-64\implies 0=(b-32)(b+2)\implies b= \begin{cases} \boxed{32}\\ -2 \end{cases}$

since the rate of speed wouldn't be a negative unit, then is not -2.