Question

Patrick started walking at a constant speed along a straight road from his school to the park. One hour after Patrick left, Tanya started running at a constant speed of $2$ miles per hour faster than Patrick walked, following the same straight road from the school to the park. One hour after Tanya left, José started bicycling at a constant speed of $7$ miles per hour faster than Tanya ran, following the same straight road from the school to the park. All three people arrived at the park at the same time. The distance from the school to the park is $\frac{m}{n}$ miles, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Answer: 277

Concepts:distance rate and time system of equations

Difficulty rating: 1840

Solution:

Let $v$ be Patrick's speed in miles per hour and $T$ his travel time in hours. Then Tanya travels for $T - 1$ hours at speed $v + 2,$ and José travels for $T - 2$ hours at speed $v + 9$ (which is $7$ more than Tanya's speed). Since all three cover the same distance, $\begin{aligned} vT &= (v+2)(T-1) \\ &= (v+9)(T-2). \end{aligned}$

Expanding the first equality gives $0 = 2T - v - 2,$ so $v = 2T - 2.$ Expanding the second gives $0 = 9T - 2v - 18,$ so $2v = 9T - 18.$ Substituting, $4T - 4 = 9T - 18,$ hence $T = \frac{14}{5}$ and $v = \frac{18}{5}.$

The distance is $vT = \frac{18}{5} \cdot \frac{14}{5} = \frac{252}{25},$ which is in lowest terms, so $m + n = 252 + 25 = 277.$

2026 AIME I Problem 1

Answer: 277

Difficulty rating: 1840

Solution:

Problem 1 in Other Years