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Difference between revisions of "2005 AMC 10A Problems/Problem 7"

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Josh and Mike live <math>13</math> miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?  
 
Josh and Mike live <math>13</math> miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?  
  
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<math> \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 </math>
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<math> \textbf{(A) } 4\qquad \textbf{(B) } 5\qquad \textbf{(C) } 6\qquad \textbf{(D) } 7\qquad \textbf{(E) } 8 </math>
  
 
==Solution==
 
==Solution==
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<math> \frac{13}{5}m = 13 </math>  
 
<math> \frac{13}{5}m = 13 </math>  
  
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<math> m = 5 \Longrightarrow \mathrm{(B)} </math>
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<math> m = \boxed{\textbf{(B) }5} </math>
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==See Also==
 
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*[[2005 AMC 10A Problems]]
 
  
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*[[2005 AMC 10A Problems/Problem 6|Previous Problem]]
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==See also==
 +
{{AMC10 box|year=2005|ab=A|num-b=6|num-a=8}}
  
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*[[2005 AMC 10A Problems/Problem 8|Next Problem]]
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{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]
 

Latest revision as of 16:28, 10 August 2023

Problem

Josh and Mike live $13$ miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?

$\textbf{(A) } 4\qquad \textbf{(B) } 5\qquad \textbf{(C) } 6\qquad \textbf{(D) } 7\qquad \textbf{(E) } 8$

Solution

Let $m$ be the distance in miles that Mike rode.

Since Josh rode for twice the length of time as Mike and at four-fifths of Mike's rate, he rode $2\cdot\frac{4}{5}\cdot m = \frac{8}{5}m$ miles.

Since their combined distance was $13$ miles,

$\frac{8}{5}m + m = 13$

$\frac{13}{5}m = 13$

$m = \boxed{\textbf{(B) }5}$

See also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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 AMC 10 Problems and Solutions

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