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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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==Solution 2==
 
==Solution 2==
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From the answer choices, the cars will meet somewhere along the <math>40</math> mph stretch. Car <math>A</math> travels <math>25</math>mph for <math>5</math> miles, so we can use dimensional analysis to see that it will be <math>\frac{1\ \text{hr}}{25\ \text{mi}}\cdot 5\ \text{mi} = \frac{1}{5}</math> of an hour for this portion. Similarly, car <math>B</math> spends <math>\frac{1}{4}</math> of an hour on the <math>20</math> mph portion. \\
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From the answer choices, the cars will meet somewhere along the <math>40</math> mph stretch. Car <math>A</math> travels <math>25</math>mph for <math>5</math> miles, so we can use dimensional analysis to see that it will be <math>\frac{1\ \text{hr}}{25\ \text{mi}}\cdot 5\ \text{mi} = \frac{1}{5}</math> of an hour for this portion. Similarly, car <math>B</math> spends <math>\frac{1}{4}</math> of an hour on the <math>20</math> mph portion.
  
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Suppose that car <math>A</math> travels <math>x</math> miles along the <math>40</math> mph portion-- then car <math>B</math> travels <math>5-x</math> miles along the <math>40</math> mph portion. By identical methods, car <math>A</math> travels for <math>\frac{1}{40}\cdot x = \frac{x}{40}</math> hours, and car <math>B</math> travels for <math>\frac{5-x}{40}</math> hours. \\
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Suppose that car <math>A</math> travels <math>x</math> miles along the <math>40</math> mph portion-- then car <math>B</math> travels <math>5-x</math> miles along the <math>40</math> mph portion. By identical methods, car <math>A</math> travels for <math>\frac{1}{40}\cdot x = \frac{x}{40}</math> hours, and car <math>B</math> travels for <math>\frac{5-x}{40}</math> hours.  
  
 
At their meeting point, cars <math>A</math> and <math>B</math> will have traveled for the same amount of time, so we have  
 
At their meeting point, cars <math>A</math> and <math>B</math> will have traveled for the same amount of time, so we have  
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<cmath>
 
\begin{align*}
 
\begin{align*}
 
     \frac{1}{5} + \frac{x}{40} &= \frac{1}{4} + \frac{5-x}{40}\\
 
     \frac{1}{5} + \frac{x}{40} &= \frac{1}{4} + \frac{5-x}{40}\\
 
     8 + x &= 10 + 5-x,
 
     8 + x &= 10 + 5-x,
 
\end{align*}
 
\end{align*}
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</cmath>
 
so <math>2x = 7</math>, and <math>x = 3.5</math> miles. This means that car <math>A</math> will have traveled <math>5 + 3.5= \boxed{\textbf{(D)\ 8.5}}</math> miles.  
 
so <math>2x = 7</math>, and <math>x = 3.5</math> miles. This means that car <math>A</math> will have traveled <math>5 + 3.5= \boxed{\textbf{(D)\ 8.5}}</math> miles.  
  
 
-Benedict T (countmath1)
 
-Benedict T (countmath1)
  
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==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
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 +
==Video Solution by Thinking Feet==
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https://youtu.be/PKMpTS6b988
 +
 +
==See Also==
 +
{{AMC8 box|year=2025|num-b=18|num-a=20}}
 +
{{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 (Problems • Answer Key • Resources)
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
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