2006 AMC 8 Problems/Problem 13

Problem

Cassie leaves Escanaba at 8:30 AM heading for Marquette on her bike. She bikes at a uniform rate of 12 miles per hour. Brian leaves Marquette at 9:00 AM heading for Escanaba on his bike. He bikes at a uniform rate of 16 miles per hour. They both bike on the same 62-mile route between Escanaba and Marquette. At what time in the morning do they meet?

$\textbf{(A)}\ 10: 00\qquad\textbf{(B)}\ 10: 15\qquad\textbf{(C)}\ 10: 30\qquad\textbf{(D)}\ 11: 00\qquad\textbf{(E)}\ 11: 30$

Solution 1

If Cassie leaves $\frac{1}{2}$ an hour earlier then Brian, when Brian starts, the distance between them will be $62-\frac{12}{2}=56$ . Every hour, they will get $12+16=28$ miles closer. $\frac{56}{28}=2$ , so 2 hours from $9:00$ AM is when they meet, which is $\boxed{\textbf{(D)}\ 11: 00}$ .

Solution 2 (Basic Algebra)

Let $x$ be the # of hours after Cassie leaves, when both of them meet. In that $x$ hours, Cassie will travel $12x$ miles. But, Brian will travel $16(x-1/2)$ miles as he starts $1/2$ hours after Cassie. These two distances sum to the total distance of $62$ miles as they meet here, yielding the equation, $12x+16(x-1/2)=62$ . After distribution $16$ we get $12x+16x-8=62$ . After combining like terms and adding $8$ to both sides we get $28x=70$ , and $x=5/2$ . So they meet $5/2$ hours after 8:30 a.m. which is $\boxed{\textbf{(D)}\ 11: 00}$ .

- LearnForEver

Video Solution

https://youtu.be/EnpR7rjMYzg Soo, DRMS, NM

Video Solution by WhyMath

https://youtu.be/IomJc_tK0x4

2006 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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