Question

A motorboat travels 104 kilometers in 4 hours going upstream. It travels 200 kilometers going downstream in the same amount of time. What is the rate of the boat in still water and what is the rate of the current?

Answer 1

$\Large \mathbb{SOLUTION:}$

$\begin{array}{l} \text{Let }r\text{ be the rate of the boat in still water and} \\ c\text{ be the rate of the current.} \\ \\ \text{So } \\ \begin{aligned} \quad&\bullet\:\text{Rate Upstream}= r - c \\ &\bullet\:\text{Rate Downstream}= r - c\end{aligned} \\ \\ \text{We know that }\text{Rate} = \dfrac{\text{Distance}}{\text{Time}}. \end{array}$

$\begin{array}{l} \bold{Equations:} \\ \\ \begin{aligned} &\quad\quad \quad r - c = \dfrac{104}{4} = 26\quad (1) \\ \\ & \quad \quad \quad r + c = \dfrac{200}{4} = 50\quad (2)\\ \\ & \text{Adding (1) and (2), we get} \\ \\ &\quad\quad 2r = 76 \implies \boxed{r = 38\ \text{kph}} \\ \\ &\text{Using (2), it follows that} \\ \\ & \quad \quad c = 50 - r \implies \boxed{c = 12\ \text{kph}} \end{aligned} \end{array}$