2025 AMC 10A Problem 1

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

The only real content is the staggered start, and the clean way to handle it is to measure time from one fixed moment and subtract the delay inside the other rider's expression. Writing $t - 1$ rather than introducing a second variable keeps the equation linear in one unknown, and the final substitution back into the clock is the only place the time of day matters.

Think of it as a catch-up problem. At $2:30$, when Betsy starts, Andy is $8$ miles from home and Betsy is $0$ miles from home. Betsy closes that gap at $12 - 8 = 4$ miles per hour, so she needs $8/4 = 2$ hours, reaching $4:30$.

• Using the right angle between the two routes and computing the distance between the riders, when the question compares each rider's distance from the common start.
• Measuring both riders' times from the same instant, which gives $8t = 12t$ and no solution.
• Solving correctly for the elapsed time $t = 3$ but adding it to Betsy's $2:30$ departure instead of Andy's $1:30$.