2005 AMC 10A Problem 7

Distance is speed times time. The problem compares both Josh's speed and Josh's time to Mike's, so compare their distances first.

1. Josh versus Mike. Josh rode for $2$ times as long, at $\frac{4}{5}$ of the speed. His distance is
$$2 \times \frac{4}{5} = \frac{8}{5}$$
times Mike's distance.
2. Think in parts. If Mike rode $5$ parts, Josh rode $8$ parts. That is the same ratio $8$ to $5$.
3. Use the total. They rode toward each other until they met, so together they covered the whole $13$ miles. That is $5 + 8 = 13$ parts, so each part is $1$ mile.
4. Mike's distance. $5$ parts, which is $5$ miles.

Check: Josh rode $8$ miles and Mike rode $5$, a total of $13$, and $8$ is $\frac{8}{5}$ of $5$.

The answer is $\boxed{\textbf{(B)}\ 5}$.

When two people ride toward each other and meet, their distances add up to the starting gap. You never need the actual speeds or times here: because the problem gives Josh's speed and time as multiples of Mike's, their product gives the ratio of distances directly, and splitting $13$ in that ratio finishes the job.

Use a letter. Call Mike's distance $m$. Josh's distance is $\frac{8}{5}m$, so $m + \frac{8}{5}m = 13$. That is $\frac{13}{5}m = 13$, so $m = 5$.

Suppose Mike rode at rate $r$ for time $t$, covering $rt$ miles. Josh rode for time $2t$ at rate $\frac{4}{5}r$, so his distance is
$$\frac{4}{5}r \cdot 2t = \frac{8}{5}\,rt.$$
Josh's distance is $\frac{8}{5}$ of Mike's. When they meet, the two distances make up the whole $13$ miles. With $m = rt$ for Mike's distance,
$$m + \frac{8}{5}m = \frac{13}{5}m = 13 \quad\Longrightarrow\quad m = 5.$$

The answer is $\boxed{\textbf{(B)}\ 5}$.