Question

A tourist sailed against the current on a river for 6 km, and then he sailed in a lake for 15 km. In the lake, he sailed for 1 hour longer than he sailed in the river. Knowing that the current of the river is 2 km/hour, find the speed of the boat while it is traveling in the lake.

Answer 1

so hmm the distances are clearly 6km and 15km anyway

so hmmm alrite.. recall your d = rt, or distance = rate * time

if the rate of speed of the tourist's boat is say "r", on the river, against the stream, is not going to be "r" but we have to subtract the river's speed, since we know is "2", thus the speed of the boat is " r - 2"

now, off he goes on the lake, however, the lake is just still water, so, on still water, his speed is indeed just "r"

if he sailed "t" time in the river, on the lake was 1 extra hour, or " t + 1 "

thus   $\bf \begin{array}{lccclll} &distance&rate&time\\ &-----&-----&-----\\ \textit{upstream at the river}&6&r-2&t\\ \textit{on the lake}&15&r&t+1 \end{array} \\\\\\ \begin{cases} 6=t(r-2)\implies \frac{6}{r-2}=\boxed{t}\\\\ 15=r(t+1)\\ ----------\\ 15=r\left( \boxed{\frac{6}{r-2}}+1 \right) \end{cases}$

solve for "r"