, 2026

Solution 1:

Each hour, Anna jogs $6\text{ km}$
and Yao jogs $8\text{ km}$, and so
combined they jog $\$6\text{ km}+8\text{ km}=14\text{}$ km}$.

After $2$ hours, they combine to jog
$2\times14\text{ km}=28\text{ km}$,
and after $3$ hours, they combine to
jog $\$3\times14\text{ km}=42\text{}$
km}$.

Since they started $42\text{ km}$
apart, then after $3$ hours they
meet. After $3$ hours, Anna has
jogged $\$3\text{ h}\times 6\text{ km/h}=18\text{}$ km}$, Yao has jogged$$3\text{ h}\times8\text{ km/h}=24\text{}$
km}$, and so Yao has travelled$$24\text{ km}-18\text{ km}=6\text{}$ km}$
farther than Anna.

Solution 2:

Anna jogs at $6\text{ km/h}$ and
Yao jogs at $8\text{ km/h}$, and so
they are moving toward one another at a constant rate of $\$6\text{ km/h}+8\text{ km/h}=14\text{}$
km/h}$.

Thus combined Anna and Yao will travel $42\text{ km}$, and meet after $\dfrac{42\text{ km}}{14\text{ km/h}}=3$
hours.

Since Yao jogs $\$8\text{ km/h}-6\text{ km/h}=2\text{}$ km/h}$ faster than Anna jogs, then Yao travels
$\$2\text{ km/h}\times 3\text{ h}=6\text{}$
km}$ farther than Anna.