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2014 AMC 10A Problem 15

2014 AMC 10A Problem 15AMC 10 Step by Step

Name: Date: October 28, 2026

Linear Equations & Word Problems · difficulty 3 of 5 · about 4 min · position P11-15

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?

A) $140$
B) $175$
C) $210$
D) $245$
E) $280$

Answer

Key insight

Look only at the distance left after the first hour. Driving it at 35 instead of 50 miles per hour takes exactly one and a half hours longer.

Solution

The first hour is the same in both plans, so only the rest of the trip matters. Call the distance still to go after the first hour $r$ miles.

1. Plan 1: keep going at $35$ mph. The rest takes $\frac{r}{35}$ hours, and he arrives $1$ hour late.
2. Plan 2: speed up to $35 + 15 = 50$ mph. The rest takes $\frac{r}{50}$ hours, and he arrives $\frac12$ hour early.
3. Compare the plans. Going from "1 hour late" to "half an hour early" is a gap of $1 + \frac12 = \frac32$ hours. So the slow plan takes $\frac32$ hours longer:
$\frac{r}{35} - \frac{r}{50} = \frac32.$
4. Clear the fractions. Multiply every term by $350$ (a multiple of $35$ , $50$ and $2$ ): $10r - 7r = 525$ , so $3r = 525$ and $r = 175$ .
5. Total distance. Add the first $35$ miles: $35 + 175 = 210$ miles.

Check: $175$ miles take $5$ hours at $35$ mph and $3.5$ hours at $50$ mph, a difference of $1.5$ hours.

The answer is $\boxed{\textbf{(C)}\ 210}$ .

Why this works

We never learn when the flight leaves, but we don't need to. Comparing the two plans cancels the unknown departure time and leaves only the time difference, $1.5$ hours. The key choice is to write the equation for the distance remaining after the first hour, since that is the only part where the two plans differ.

Alternative approach

Test the choices. For a total distance $d$ , the remaining $d - 35$ miles must take $1.5$ hours longer at $35$ mph than at $50$ mph.

total	remaining	time at 35	time at 50	difference
$175$	$140$	$4$	$2.8$	$1.2$
$210$	$175$	$5$	$3.5$	$1.5$

Only $210$ works.

The trap

Treating 1 hour late and 30 minutes early as a 30-minute difference, or using the faster speed for the first hour too.

Concise solution

After the first hour, David still has $r$ miles to go. Everything that follows concerns only this remaining stretch.

Continuing at $35$ mph would take $\frac{r}{35}$ hours and make him $1$ hour late.
Driving at $35 + 15 = 50$ mph takes $\frac{r}{50}$ hours and makes him $\frac12$ hour early.

The difference between these two arrival times is $1 + \frac12 = \frac32$ hours, so
$\frac{r}{35} - \frac{r}{50} = \frac32.$
Multiply by $350$ : $10r - 7r = 525$ , giving $3r = 525$ and $r = 175$ .

The total distance is the first $35$ miles plus the remaining $175$ :
$35 + 175 = 210.$

The answer is $\boxed{\textbf{(C)}\ 210}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 7: Proportion (§7.4)

Techniques

Set up the equation/formula and compute; no special trick needed