CEMC Fermat, 2014

As Jillian drives from $J$ to $G$, suppose that she drives $x$ km uphill, $y$ km on flat ground, and $z$ km downhill.

This means that when she drives from $G$ to $J$, she will drive $z$ km uphill, $y$ km on flat ground, and $x$ 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 km/h on flat ground, 63 km/h uphill, and 99 km/h downhill.

Since time equals distance divided by speed, then on her trip from $J$ to $G$, her time driving uphill is $\dfrac{x}{63}$ hours, her time driving on flat ground is $\dfrac{y}{77}$ hours, and her time driving downhill is $\dfrac{z}{99}$ hours.

Since it takes her 3 hours and 40 minutes (which is $3\dfrac{2}{3}$ or $\dfrac{11}{3}$ hours), then $$\dfrac{x}{63}+\dfrac{y}{77} + \dfrac{z}{99} = \dfrac{11}{3}$$ A similar analysis of the return trip gives $$\dfrac{x}{99}+\dfrac{y}{77} + \dfrac{z}{63} = \dfrac{13}{3}$$ We are asked for the total distance from $J$ to $G$, which equals $x+y+z$ km. Therefore, we need to determine $x+y+z$.

We add the two equations above and simplify to obtain $$\begin{aligned} \dfrac{x}{63}+\dfrac{x}{99} + \dfrac{y}{77}+\dfrac{y}{77} + \dfrac{z}{99}+\dfrac{z}{63} &= \dfrac{24}{3}\\ x\left(\dfrac{1}{63}+\dfrac{1}{99}\right) + y\left(\dfrac{1}{77}+\dfrac{1}{77}\right) + z\left(\dfrac{1}{99}+\dfrac{1}{63}\right) & = 8 \\ x\left(\dfrac{1}{7 \cdot 9}+\dfrac{1}{9 \cdot 11}\right) + \dfrac{2}{77}y + z\left(\dfrac{1}{9\cdot 11}+\dfrac{1}{7 \cdot 9}\right) & = 8 \\ x\left(\dfrac{11}{7 \cdot 9 \cdot 11}+\dfrac{7}{7 \cdot 9 \cdot 11}\right) + \dfrac{2}{77}y + z\left(\dfrac{7}{7 \cdot 9 \cdot 11}+\dfrac{11}{7 \cdot 9 \cdot 11}\right) & = 8 \\ x\left(\dfrac{18}{7 \cdot 9 \cdot 11}\right) + \dfrac{2}{77}y + z\left(\dfrac{18}{7 \cdot 9 \cdot 11}\right) & = 8 \\ x\left(\dfrac{2}{7 \cdot 11}\right) + \dfrac{2}{77}y + z\left(\dfrac{2}{7 \cdot 11}\right) & = 8 \\ \dfrac{2}{77}(x+y+z) & = 8\end{aligned}$$ Thus, $x+y+z = \dfrac{77}{2}\cdot 8 = 77\cdot 4 = 308$.

Finally, the distance from $J$ to $G$ is 308 km.