2014 AMC 10A Problem 15

Below is the professionally curated solution for Problem 15 of the 2014 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AMC 10A solutions, or check the answer key.

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Concepts:distance rate and timelinear equation

Difficulty rating: 1540

15.

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?

140140

175175

210210

245245

280280

Solution:

Note that David drives at 5050 miles per hour after one hour. Let the distance he still needs to be drive be d.d.

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{align*} \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{align*} Therefore, the airport is x=210x=210 miles from David's house.

Thus, C is the correct answer.

Problem 15 in Other Years