Two cyclists, $k$ kilometres apart, and starting at the same time, would be together in $r$ hours if they travelled in the same direction, but would pass each other in $t$ hours if they travelled in opposite directions. The ratio of the speed of the faster cyclist to that of the slower is

### Concept & Relative Speed Variables Using generalized variables to create a ratio requires translating "same direction" (difference of speeds) and "opposite directions" (sum of speeds) into linear equations. $$Speed = \frac{Distance}{Time}$$ ### Step-by-Step Solution 1. Let the speed of the faster cyclist be $u$ and the slower cyclist be $v$. The distance is $k$. 2. **Same direction (together in $r$ hours):** Relative speed is $(u - v)$. $$(u - v) = \frac{k}{r}$$ 3. **Opposite direction (pass in $t$ hours):** Relative speed is $(u + v)$. $$(u + v) = \frac{k}{t}$$ 4. **Find the ratio $\frac{u}{v}$:** Divide the second equation by the first equation. $$\frac{u + v}{u - v} = \frac{\frac{k}{t}}{\frac{k}{r}}$$ $$\frac{u + v}{u - v} = \frac{r}{t}$$ 5. Apply *Componendo and Dividendo* (if $\frac{a}{b} = \frac{c}{d}$, then $\frac{a+b}{a-b} = \frac{c+d}{c-d}$): $$\frac{(u+v) + (u-v)}{(u+v) - (u-v)} = \frac{r + t}{r - t}$$ $$\frac{2u}{2v} = \frac{r + t}{r - t}$$ $$\frac{u}{v} = \frac{r + t}{r - t}$$ ### Exam Strategy & Shortcut Without writing equations, realize that $\text{Ratio of Speeds} = \frac{\text{Sum of Times}}{\text{Difference of Times}}$. Because opposite direction takes less time ($t$) and same direction takes more time ($r$), the ratio of faster to slower speed is strictly $\frac{r+t}{r-t}$. ### Common Pitfall Setting the same direction relative speed to $(u+v)$ instead of $(u-v)$. Same direction means the faster object must catch up, so relative speed is their difference. ### Final Answer Therefore, the correct answer is **$\frac{r + t}{r - t}$**.