Difference between revisions of "2025 AMC 8 Problems/Problem 19"

 
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==Problem==
 
Two towns, <math>A</math> and <math>B</math>, are connected by a straight road, <math>15</math> miles long. Traveling from town <math>A</math> to town <math>B</math>, the speed limit changes every <math>5</math> miles: from <math>25</math> to <math>40</math> to <math>20</math> miles per hour (mph). Two cars, one at town <math>A</math> and one at town <math>B</math>, 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 <math>A</math>, in miles, will the two cars meet?
 
Two towns, <math>A</math> and <math>B</math>, are connected by a straight road, <math>15</math> miles long. Traveling from town <math>A</math> to town <math>B</math>, the speed limit changes every <math>5</math> miles: from <math>25</math> to <math>40</math> to <math>20</math> miles per hour (mph). Two cars, one at town <math>A</math> and one at town <math>B</math>, 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 <math>A</math>, in miles, will the two cars meet?
  
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-Benedict T (countmath1)
 
-Benedict T (countmath1)
  
==Vide Solution 1 by SpreadTheMathLove==
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==Video Solution 1 by SpreadTheMathLove==
 
https://www.youtube.com/watch?v=jTTcscvcQmI
 
https://www.youtube.com/watch?v=jTTcscvcQmI
  
 
==Video Solution by Thinking Feet==
 
==Video Solution by Thinking Feet==
 
https://youtu.be/PKMpTS6b988
 
https://youtu.be/PKMpTS6b988
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==See Also==
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{{AMC8 box|year=2025|num-b=18|num-a=20}}
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{{MAA Notice}}

Latest revision as of 19:10, 30 January 2025

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?

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

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 1 by SpreadTheMathLove

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

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

See Also

2025 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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