# 2014 AMC 10A Problems/Problem 15

The following problem is from both the 2014 AMC 12A #11 and 2014 AMC 10A #15, so both problems redirect to this page.

## Problem

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?

$\textbf{(A) }140\qquad \textbf{(B) }175\qquad \textbf{(C) }210\qquad \textbf{(D) }245\qquad \textbf{(E) }280\qquad$

## Solution 1

Note that he drives at $50$ miles per hour after the first hour and continues doing so until he arrives.

Let $d$ be the distance still needed to travel after $1$ hour. We have that $\dfrac{d}{50}+1.5=\dfrac{d}{35}$, where the $1.5$ comes from $1$ hour late decreased to $0.5$ hours early.

Simplifying gives $7d+525=10d$, or $d=175$.

Now, we must add an extra $35$ miles traveled in the first hour, giving a total of $\boxed{\textbf{(C) } 210}$ miles.

Instead of spending time thinking about how one can set up an equation to solve the problem, one can simply start checking the answer choices. Quickly checking, we know that neither choice $\textbf{(A)}$ or choice $\textbf{(B)}$ work, but $\textbf{(C)}$ does. We can verify as follows. After $1$ hour at $35 \text{ mph}$, David has $175$ miles left. This then takes him $3.5$ hours at $50 \text{ mph}$. But $210/35 = 6 \text{ hours}$. Since $1+3.5 = 4.5 \text{ hours}$ is $1.5$ hours less than $6$, our answer is $\boxed{\textbf{(C)} \: 210}$.

## Solution 3 (similar to Solution 1 but using a different equation)

Let the total distance be $d$. Then $d=35(t+1)$. Since 1 hour has passed, and he increased his speed by $15$ miles per hour, and he had already traveled $35$ miles, the new equation is $d-35=50(t-1-\frac{1}{2})=50(t-\frac{3}{2})$. Solving, $d=35t+35=50t-40$, $15t=75$, $t=5$, and $d=35(5+1)=35\cdot 6=210 \Longrightarrow \boxed{\textbf{(C) } 210}$.

~JH. L

 2014 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byProblem 16 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
 2014 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions