Jillian drives along a straight road that goes directly from her house $(J)$ to her Grandfather's house $(G)$. Some of this road is on flat ground and some is downhill or uphill. Her car travels downhill at $99 \mathrm{~km} / \mathrm{h}$, on flat ground at $77 \mathrm{~km} / \mathrm{h}$, and uphill at $63 \mathrm{~km} / \mathrm{h}$. It takes Jillian 3 hours and 40 minutes to drive from $J$ to $G$. It takes her 4 hours and 20 minutes to drive from $G$ to $J$. The distance between $J$ and $G$, in $\mathrm{km}$, is
(A) $318 \frac{2}{3}$
(B) 324
(C) 308
(D) $292 \frac{3}{5}$
(E) $292 \frac{1}{9}$
As Jillian drives from $J$ to $G$, suppose that she drives $x \mathrm{~km}$ uphill, $y \mathrm{~km}$ on flat ground, and $z \mathrm{~km}$ downhill.

This means that when she drives from $G$ to $J$, she will drive $z \mathrm{~km}$ uphill, $y \mathrm{~km}$ on flat ground, and $x \mathrm{~km}$ downhill. This is because downhill portions become uphill portions on the return trip, while uphill portions become downhill portions on the return trip.

We are told that Jillian drives at $77 \mathrm{~km} / \mathrm{h}$ on flat ground, $63 \mathrm{~km} / \mathrm{h}$ uphill, and $99 \mathrm{~km} / \mathrm{h}$ downhill. Since time equals distance divided by speed, then on her trip from $J$ to $G$, her time driving uphill is $\frac{x}{63}$ hours, her time driving on flat ground is $\frac{y}{77}$ hours, and her time driving downhill is $\frac{z}{99}$ hours.

Since it takes her 3 hours and 40 minutes (which is $3 \frac{2}{3}$ or $\frac{11}{3}$ hours), then

$$
\frac{x}{63}+\frac{y}{77}+\frac{z}{99}=\frac{11}{3}
$$

A similar analysis of the return trip gives

$$
\frac{x}{99}+\frac{y}{77}+\frac{z}{63}=\frac{13}{3}
$$

We are asked for the total distance from $J$ to $G$, which equals $x+y+z \mathrm{~km}$. Therefore, we need to determine $x+y+z$.

We add the two equations above and simplify to obtain

$$
\begin{aligned}
\frac{x}{63}+\frac{x}{99}+\frac{y}{77}+\frac{y}{77}+\frac{z}{99}+\frac{z}{63} & =\frac{24}{3} \\
x\left(\frac{1}{63}+\frac{1}{99}\right)+y\left(\frac{1}{77}+\frac{1}{77}\right)+z\left(\frac{1}{99}+\frac{1}{63}\right) & =8 \\
x\left(\frac{1}{7 \cdot 9}+\frac{1}{9 \cdot 11}\right)+\frac{2}{77} y+z\left(\frac{1}{9 \cdot 11}+\frac{1}{7 \cdot 9}\right) & =8 \\
x\left(\frac{11}{7 \cdot 9 \cdot 11}+\frac{7}{7 \cdot 9 \cdot 11}\right)+\frac{2}{77} y+z\left(\frac{7}{7 \cdot 9 \cdot 11}+\frac{11}{7 \cdot 9 \cdot 11}\right) & =8 \\
x\left(\frac{18}{7 \cdot 9 \cdot 11}\right)+\frac{2}{77} y+z\left(\frac{18}{7 \cdot 9 \cdot 11}\right) & =8 \\
x\left(\frac{2}{7 \cdot 11}\right)+\frac{2}{77} y+z\left(\frac{2}{7 \cdot 11}\right) & =8 \\
\frac{2}{77}(x+y+z) & =8
\end{aligned}
$$

Thus, $x+y+z=\frac{77}{2} \cdot 8=77 \cdot 4=308$.

Finally, the distance from $J$ to $G$ is $308 \mathrm{~km}$.