2025 AMC 8 Problems/Problem 19

Problem

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?

$[asy] // Asymptote code by aoum size(10cm); real h = 0.1; real s = 0.07; path b = brace((1,0),(0,0),amplitude=s); filldraw((0,0)--(3,0)--(3,h)--(0,h)--cycle,lightgray,black+1bp); draw((1,0)--(1,h),dashed); draw((2,0)--(2,h),dashed); label("$A$",(0,h/2),W); label("$B$",(3,h/2),E); draw(scale(0.7)*"$25\,\textrm{mph}$",(1,h+s)--(0,h+s),Bars); draw(scale(0.7)*"$40\,\textrm{mph}$",(2,h+s)--(1,h+s),Bars); draw(scale(0.7)*"$20\,\textrm{mph}$",(3,h+s)--(2,h+s),Bars); draw(b); draw(shift(1,0)*b); draw(shift(2,0)*b); label(scale(0.7)*"$5\,\textrm{mi}$",(0.5,-s),S); label(scale(0.7)*"$5\,\textrm{mi}$",(1.5,-s),S); label(scale(0.7)*"$5\,\textrm{mi}$",(2.5,-s),S); [/asy]$

$\textbf{(A)}\ 7.75\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 8.25\qquad \textbf{(D)}\ 8.5\qquad \textbf{(E)}\ 8.75$

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 $A$ .

~alwaysgonnagiveyouup

Solution 2

From the answer choices, the cars will meet somewhere along the $40$ mph stretch. Car $A$ travels $25$ mph 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.

-Benedict T (countmath1)

Video Solution by Pi Academy

https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jTTcscvcQmI

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/VP7g-s8akMY?si=Y7swThPvf2WCCGxM&t=2394 ~hsnacademy

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AJHSME/AMC 8 Problems and Solutions

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