2007 AMC 10A Problem 13

Only the ratio of distances matters, so pick easy speeds: walking speed $1$, biking speed $7$. Then time = distance $\div$ speed.

1. Name the distances. Yan is $a$ from home and $b$ from the stadium. Home to stadium is $a + b$.
2. Choice 1: walk to the stadium. Distance $b$ at speed $1$: time $b$.
3. Choice 2: walk home, then bike. Walking home takes $a$. Then he bikes the whole way from home to the stadium, distance $a + b$ at speed $7$: time $\frac{a + b}{7}$. Total: $a + \frac{a+b}{7}$.
4. The times are equal.
$$b = a + \frac{a + b}{7}.$$
5. Clear the fraction by multiplying both sides by $7$:
$$7b = 7a + a + b, \qquad 6b = 8a.$$
6. Ratio. $\frac{a}{b} = \frac{6}{8} = \frac{3}{4}$.
7. Check with numbers. Take $a = 3$, $b = 4$. Walking straight: $4$. Walking home ($3$) then biking $7$ at speed $7$ ($1$): total $4$. Equal.

The answer is $\boxed{\textbf{(B)}\ \frac 34}$.

When a problem only asks for a ratio, you may choose a convenient speed, like $1$, without changing the answer. Writing the time for each route and setting them equal gives one equation. Then read off the ratio of the two distances; you never need the actual distances.