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

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 the first $1$ 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 $\textbf{(C) }210$ miles.

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2014 AMC 10A Problems/Problem 15

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