Difference between revisions of "2006 AMC 8 Problems/Problem 13"

m
Line 10: Line 10:
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2006|num-b=12|num-a=14}}
 
{{AMC8 box|year=2006|num-b=12|num-a=14}}
 +
{{MAA Notice}}

Revision as of 01:22, 5 July 2013

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

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 $\boxed{\textbf{(D)}\ 11: 00}$.

See Also

2006 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png