2014 AMC 10A Problem 15

After the first hour, David still has $r$ miles to go. Everything that follows concerns only this remaining stretch.

• Continuing at $35$ mph would take $\frac{r}{35}$ hours and make him $1$ hour late.
• Driving at $35 + 15 = 50$ mph takes $\frac{r}{50}$ hours and makes him $\frac12$ hour early.

The difference between these two arrival times is $1 + \frac12 = \frac32$ hours, so
$$\frac{r}{35} - \frac{r}{50} = \frac32.$$
Multiply by $350$: $10r - 7r = 525$, giving $3r = 525$ and $r = 175$.

The total distance is the first $35$ miles plus the remaining $175$:
$$35 + 175 = 210.$$

The answer is $\boxed{\textbf{(C)}\ 210}$.

The scheduled arrival time is unknown, but it cancels when you compare the two scenarios: "1 hour late" versus "30 minutes early" means the faster plan saves exactly $1.5$ hours on the same remaining distance. Setting up the equation on the remaining leg, not the whole trip, is the crucial modeling choice.

Test choices: with $210$ miles, the remaining $175$ miles take $5$ hours at $35$ mph or $3.5$ hours at $50$ mph, a difference of $1.5$ hours as required. With $175$ miles the remaining $140$ miles give $4 - 2.8 = 1.2$ hours, too small.

• Applying the speed increase to the whole trip (including the first hour) or mixing up the 1 hour late and 30 minutes early as a 1-hour difference.
• Solving for the remaining distance $175$ and forgetting to add the first $35$ miles, choosing (B).