Answer
$t=\frac{10}{3r}$
Work Step by Step
Using $d=rt$ where $d$ is distance, $r$ is rate, and $t$ is time, then the equation that represents as the runner goes up a slope is
$$\begin{aligned}
2&=rt_1
\\
t_1&=\frac{2}{r}
,\end{aligned}$$
while the equation that represents as the runner goes down a slope is
$$\begin{aligned}
2&=(r+0.50r)t_2
\\
2&=1.50rt_2
\\
t_2&=\frac{2}{1.50r}
\\&=
\frac{20}{15r}
\\&=
\frac{4}{3r}
.\end{aligned}$$
Adding the two time expressions, then the time, $t$ spent by the runner is
$$\begin{aligned}
t&=t_1+t_2
\\&=
\frac{2}{r}+\frac{4}{3r}
\\&=
\frac{2}{r}\cdot\frac{3}{3}+\frac{4}{3r}
\\&=
\frac{6}{3r}+\frac{4}{3r}
\\&=
\frac{6+4}{3r}
\\&=
\frac{10}{3r}
.\end{aligned}$$Hence, the time spent by the runner is $t=\frac{10}{3r}$.
