Difference between revisions of "2006 AMC 8 Problems/Problem 13"
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If Cassie leaves <math> \frac{1}{2} </math> an hour earlier then Brian, when Brian starts, the distance between them will be <math> 62-\frac{12}{2}=56 </math>. Every hour, they will get <math> 12+16=28 </math> miles closer. <math> \frac{56}{28}=2 </math>, so 2 hours from 9:00 AM is when they meet, which is <math> \boxed{\textbf{(D)}\ 11: 00} </math>. | If Cassie leaves <math> \frac{1}{2} </math> an hour earlier then Brian, when Brian starts, the distance between them will be <math> 62-\frac{12}{2}=56 </math>. Every hour, they will get <math> 12+16=28 </math> miles closer. <math> \frac{56}{28}=2 </math>, so 2 hours from 9:00 AM is when they meet, which is <math> \boxed{\textbf{(D)}\ 11: 00} </math>. | ||
+ | ==See Also== | ||
{{AMC8 box|year=2006|num-b=12|num-a=14}} | {{AMC8 box|year=2006|num-b=12|num-a=14}} |
Revision as of 20:03, 24 December 2012
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?
Solution
If Cassie leaves an hour earlier then Brian, when Brian starts, the distance between them will be . Every hour, they will get miles closer. , so 2 hours from 9:00 AM is when they meet, which is .
See Also
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 |