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

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<math>\textbf{(A)}\ 7.75\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 8.25\qquad \textbf{(D)}\ 8.5\qquad \textbf{(E)}\ 8.75</math>
 
<math>\textbf{(A)}\ 7.75\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 8.25\qquad \textbf{(D)}\ 8.5\qquad \textbf{(E)}\ 8.75</math>
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==Solution 1==
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The first car, moving from town <math>A</math> at <math>25</math> miles per hour, takes <math>\frac{5}{25} = \frac{1}{5} \text{hours} = 12</math> minutes. The second car, traveling another <math>5</math> miles from town <math>B</math>, takes <math>\frac{5}{20} = \frac{1}{4} \text{hours} = 15</math> minutes. The first car has traveled for 3 minutes or <math>\frac{1}{20}</math>th of an hour at <math>40</math> miles per hour when the second car has traveled 5 miles. The first car has traveled <math>40 \cdot \frac{1}{20} = 2</math> miles from the previous <math>5</math> miles it traveled at <math>25</math> miles per hour. They have <math>3</math> miles left, and they travel at the same speed, so they meet <math>1.5</math> miles through, so they are <math>5 + 2 + 1.5 = \boxed{\textbf{(D)}8.5}</math> miles from town <math>A</math>.
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~alwaysgonnagiveyouup

Revision as of 22:07, 29 January 2025

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