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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 <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)==
 
==Solution 2 (Basic Algebra)==
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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>.
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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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- LearnForTheWin
+
- LearnForEver
  
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/EnpR7rjMYzg Soo, DRMS, NM
 
https://youtu.be/EnpR7rjMYzg Soo, DRMS, NM
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 +
==Video Solution by WhyMath==
 +
https://youtu.be/IomJc_tK0x4
  
 
==See Also==
 
==See Also==

Latest revision as of 15:13, 30 October 2024

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

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

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