2025 AMC 8

Complete problem set with solutions and individual problem pages

Problem 19 Hard

Two towns, A and B, are connected by a straight road, 15 miles long. Traveling from town A to town B, the speed limit changes every 5 miles: from 25 to 40 to 20 miles per hour (mph). Two cars, one at town A and one at town B, start moving toward each other at the same time. They drive at exactly the speed limit in each portion of the road. How far from town A, in miles, will the two cars meet?

  • A.

    7.75

  • B.

    8

  • C.

    8.25

  • D.

    8.5

  • E.

    8.75

Answer:D

Solution 1

The first car, moving from town A at 25 miles per hour, takes \frac{5}{25} = \frac{1}{5} \text{hours} = 12 minutes. The second car, traveling another 5 miles from town B, takes \frac{5}{20} = \frac{1}{4} \text{hours} = 15 minutes. The first car has traveled for 3 minutes or \frac{1}{20}th of an hour at 40 miles per hour when the second car has traveled 5 miles. The first car has traveled 40 \cdot \frac{1}{20} = 2 miles from the previous 5 miles it traveled at 25 miles per hour. They have 3 miles left, and they travel at the same speed, so they meet 1.5 miles through, so they are 5 + 2 + 1.5 = \boxed{\textbf{(D) }8.5} miles from town A.

 

Solution 2

From the answer choices, the cars will meet somewhere along the 40 mph stretch. Car A travels 25mph for 5 miles, so we can use dimensional analysis to see that it will be \frac{1\ \text{hr}}{25\ \text{mi}}\cdot 5\ \text{mi} = \frac{1}{5} of an hour for this portion. Similarly, car B spends \frac{1}{4} of an hour on the 20 mph portion.

Suppose that car A travels x miles along the 40 mph portion-- then car B travels 5-x miles along the 40 mph portion. By identical methods, car A travels for \frac{1}{40}\cdot x = \frac{x}{40} hours, and car B travels for \frac{5-x}{40} hours.

At their meeting point, cars A and B will have traveled for the same amount of time, so we have

\begin{align*} \frac{1}{5} + \frac{x}{40} &= \frac{1}{4} + \frac{5-x}{40}\\ 8 + x &= 10 + 5-x, \end{align*}

so 2x = 7, and x = 3.5 miles. This means that car A will have traveled 5 + 3.5= \boxed{\textbf{(D) 8.5}} miles.