2014 AMC 10A Problem 15

The first hour is the same in both plans, so only the rest of the trip matters. Call the distance still to go after the first hour $r$ miles.

1. Plan 1: keep going at $35$ mph. The rest takes $\frac{r}{35}$ hours, and he arrives $1$ hour late.
2. Plan 2: speed up to $35 + 15 = 50$ mph. The rest takes $\frac{r}{50}$ hours, and he arrives $\frac12$ hour early.
3. Compare the plans. Going from "1 hour late" to "half an hour early" is a gap of $1 + \frac12 = \frac32$ hours. So the slow plan takes $\frac32$ hours longer:
$$\frac{r}{35} - \frac{r}{50} = \frac32.$$
4. Clear the fractions. Multiply every term by $350$ (a multiple of $35$, $50$ and $2$): $10r - 7r = 525$, so $3r = 525$ and $r = 175$.
5. Total distance. Add the first $35$ miles: $35 + 175 = 210$ miles.

Check: $175$ miles take $5$ hours at $35$ mph and $3.5$ hours at $50$ mph, a difference of $1.5$ hours.

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

We never learn when the flight leaves, but we don't need to. Comparing the two plans cancels the unknown departure time and leaves only the time difference, $1.5$ hours. The key choice is to write the equation for the distance remaining after the first hour, since that is the only part where the two plans differ.

Test the choices. For a total distance $d$, the remaining $d - 35$ miles must take $1.5$ hours longer at $35$ mph than at $50$ mph.

Only $210$ works.