2014 AMC 10A Problem 15

Question

David drives from his home to the airport to catch a flight. He drives

35

miles in the first hour, but realizes that he will be

1

hour late if he continues at this speed. He increases his speed by

15

miles per hour for the rest of the way to the airport and arrives

30

minutes early. How many miles is the airport from his home?

140

175

210

245

280

Compare the original too-slow schedule with the faster schedule

The faster trip saves

1.5

hours overall

Note that David drives at

50

miles per hour after one hour.

Then, if the airport is

x

miles from David’s house, we know that:

\dfrac x{35} - \left(1+\dfrac{x-35}{50}\right) = \dfrac 32

We solve this equation as follows:

\begin{aligned} \dfrac x{35} - \left(1+\dfrac{x-35}{50}\right) &= \dfrac 32\\ \dfrac x{35} - \dfrac{x-35+50}{50} &= \dfrac 32\\ 10x - 7(x+15)&= 525\\ 10x - 7x-105&= 525\\ 3x&= 630\\ x&=210 \end{aligned}

Therefore, the airport is

x=210

miles from David’s house.

Thus, C is the correct answer.

Problem 15 in other years