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 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 <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 two 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>. | ||
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| + | - LearnForEver | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/EnpR7rjMYzg Soo, DRMS, NM | ||
==See Also== | ==See Also== | ||
Revision as of 15:48, 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 two 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 .
- LearnForEver
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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