Difference between revisions of "2006 AMC 8 Problems/Problem 13"
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<math> \textbf{(A)}\ 10: 00\qquad\textbf{(B)}\ 10: 15\qquad\textbf{(C)}\ 10: 30\qquad\textbf{(D)}\ 11: 00\qquad\textbf{(E)}\ 11: 30 </math> | <math> \textbf{(A)}\ 10: 00\qquad\textbf{(B)}\ 10: 15\qquad\textbf{(C)}\ 10: 30\qquad\textbf{(D)}\ 11: 00\qquad\textbf{(E)}\ 11: 30 </math> | ||
− | == Solution == | + | == Solution 1== |
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 <math>9:00</math> 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 <math>9:00</math> AM is when they meet, which is <math> \boxed{\textbf{(D)}\ 11: 00} </math>. | ||
+ | ==Solution 2 (Basic Algebra)== | ||
+ | Let <math>x</math> be the # of hours after Cassie leaves, when both of them meet. In that <math>x</math> hours, Cassie will travel <math>12x</math> miles. But, Brian will travel <math>16(x-1/2)</math> miles as he starts <math>1/2</math> hours after Cassie. These to distances sum to the total distance of <math>62</math> miles as they meet here, yielding the equation, <math>12x+16(x-1/2)=62</math>. After distribution <math>16</math> we get <math>12x+16x-8=62</math>. After combining like terms and adding <math>8</math> to both sides we get <math>28x=70</math>, and <math>x=5/2</math>. So they meet <math>5/2</math> hours after 8:30 a.m. which is <math>\boxed{\textbf{(D)}\ 11: 00}</math>. | ||
+ | -LearnForTheWin | ||
==Video Solution== | ==Video Solution== |
Revision as of 16:45, 29 October 2023
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 1
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
AM is when they meet, which is
.
Solution 2 (Basic Algebra)
Let be the # of hours after Cassie leaves, when both of them meet. In that
hours, Cassie will travel
miles. But, Brian will travel
miles as he starts
hours after Cassie. These to distances sum to the total distance of
miles as they meet here, yielding the equation,
. After distribution
we get
. After combining like terms and adding
to both sides we get
, and
. So they meet
hours after 8:30 a.m. which is
.
-LearnForTheWin
Video Solution
https://youtu.be/EnpR7rjMYzg Soo, DRMS, NM
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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