Skip to main content

2014 AMC 10A Problem 15

Problem 15 of 25IntermediateAlgebra

David drives from his home to the airport to catch a flight. He drives 3535 miles in the first hour, but realizes that he will be 11 hour late if he continues at this speed. He increases his speed by 1515 miles per hour for the rest of the way to the airport and arrives 3030 minutes early. How many miles is the airport from his home?

Answer choices

Show solution

Solution

Note that David drives at 5050 miles per hour after one hour. Then, if the airport is xx miles from David’s house, we know that: x35(1+x3550)=32\dfrac x{35} - \left(1+\dfrac{x-35}{50}\right) = \dfrac 32 We solve this equation as follows: x35(1+x3550)=32x35x35+5050=3210x7(x+15)=52510x7x105=5253x=630x=210\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=210x=210 miles from David’s house. Thus, C is the correct answer.

More practice

Concepts: distance rate and time · linear equation

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.