Suppose that the length of the route is $d \mathrm{~km}$.

Then Jill jogs $\frac{d}{2} \mathrm{~km}$ at $6 \mathrm{~km} / \mathrm{h}$ and runs $\frac{d}{2} \mathrm{~km}$ at $12 \mathrm{~km} / \mathrm{h}$.

Note that time equals distance divided by speed.

Since her total time was $x$ hours, then $x=\frac{d / 2}{6}+\frac{d / 2}{12}=\frac{d}{12}+\frac{d}{24}=\frac{2 d}{24}+\frac{d}{24}=\frac{3 d}{24}=\frac{d}{8}$.

Also, Jack walks $\frac{d}{3} \mathrm{~km}$ at $5 \mathrm{~km} / \mathrm{h}$ and runs $\frac{2 d}{3} \mathrm{~km}$ at $15 \mathrm{~km} / \mathrm{h}$.

Since his total time is $y$ hours, then $y=\frac{d / 3}{5}+\frac{2 d / 3}{15}=\frac{d}{15}+\frac{2 d}{45}=\frac{3 d}{45}+\frac{2 d}{45}=\frac{5 d}{45}=\frac{d}{9}$.

Finally, $\frac{x}{y}=\frac{d / 8}{d / 9}=\frac{9}{8}$.

Answer: (A)
Jack and Jill exercise along the same route. Jill jogs the first half of the route at $6 \mathrm{~km} / \mathrm{h}$, runs the remainder of the route at $12 \mathrm{~km} / \mathrm{h}$ and takes a total time of $x$ hours. Jack walks the first third of the route at $5 \mathrm{~km} / \mathrm{h}$, runs the remainder at $15 \mathrm{~km} / \mathrm{h}$ and takes a total time of $y$ hours. Which of the following is equal to $\frac{x}{y}$ ?
(A) $\frac{9}{8}$
(B) $\frac{7}{5}$
(C) $\frac{15}{16}$
(D) $\frac{9}{16}$
(E) $\frac{10}{9}$

Part C: Each correct answer is worth 8.