2025 AMC 10A Problem 1

P1-10~2 minPrint

Linear Equations & Word Problems Clocks, Calendars & Time Set up the equation/formula and compute; no special trick needed

Andy and Betsy both live in Mathville. Andy leaves Mathville on his bicycle at $1:30$ , traveling due north at a steady $8$ miles per hour. Betsy leaves on her bicycle from the same point at $2:30$ , traveling due east at a steady $12$ miles per hour. At what time will they be exactly the same distance from their common starting point?

Show solution without answering

Key insight

Only each rider's distance from home matters, and that is speed times riding time. Remember Betsy has ridden one hour less than Andy.

Solution

First notice what the question asks. It compares how far each rider is from home, not how far apart they are. So the directions north and east do not matter at all.

Try a few times to get a feel. At $2:30$ , Andy has ridden $1$ hour, so he is $8$ miles out, and Betsy is just starting at $0$ miles. At $3:30$ , Andy is at $16$ miles and Betsy at $12$ . Betsy is catching up.

Time	Andy	Betsy
$2:30$	$8$	$0$
$3:30$	$16$	$12$
$4:30$	$24$	$24$

To be sure, write it as an equation. Let $t$ be the number of hours after $1:30$ . Andy has ridden $t$ hours, so he is $8t$ miles out.

Betsy left one hour later, so she has ridden only $t - 1$ hours. She is $12(t-1)$ miles out.

5. Set the two distances equal:
$8t = 12(t-1) \quad\Longrightarrow\quad 8t = 12t - 12 \quad\Longrightarrow\quad 12 = 4t \quad\Longrightarrow\quad t = 3.$

Three hours after $1:30$ is $4:30$ . Check: Andy has ridden $3$ hours, $8 \times 3 = 24$ miles; Betsy has ridden $2$ hours, $12 \times 2 = 24$ miles. They match.

The answer is $\boxed{\textbf{(E)}\ 4:30}$ .

Why this works

When two people start at different times, measure all time from one starting moment and subtract the delay for the later person, here $t - 1$ . That keeps just one unknown. Also read the question carefully: "distance from the starting point" is about each person alone, so the directions are just decoration.

Alternative approach

Think of it as a chase. When Betsy starts at $2:30$ , Andy is already $8$ miles from home. Each hour Betsy goes $12$ miles and Andy goes $8$ , so Betsy gains $12 - 8 = 4$ miles per hour on him. To make up $8$ miles she needs $8 \div 4 = 2$ hours. Two hours after $2:30$ is $4:30$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Both riders travel in a straight line away from Mathville, so each one's distance from the starting point is simply speed times time. The compass directions play no part.

Let $t$ be the number of hours after $1:30$ . Andy has ridden $8t$ miles. Betsy set off an hour later, so she has been riding for $t - 1$ hours and has covered $12(t-1)$ miles. Setting the distances equal,
$8t = 12(t-1) \implies 8t = 12t - 12 \implies 4t = 12 \implies t = 3.$
Three hours after $1:30$ is $4:30$ . As a check, Andy has then ridden $8 \cdot 3 = 24$ miles and Betsy $12 \cdot 2 = 24$ miles.

The answer is $\boxed{\textbf{(E)}\ 4:30}$ .

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)