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

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<math> \frac{13}{5}m = 13 </math>  
 
<math> \frac{13}{5}m = 13 </math>  
  
<math> m = 5 \Rightarrow B </math>
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<math> m = 5 \Longrightarrow \mathrm{(B)} </math>
 
   
 
   
 
==See Also==
 
==See Also==
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*[[2005 AMC 10A Problems/Problem 8|Next Problem]]
 
*[[2005 AMC 10A Problems/Problem 8|Next Problem]]
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[[Category:Introductory Algebra Problems]]

Revision as of 10:41, 2 August 2006

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?

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

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,

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

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

$m = 5 \Longrightarrow \mathrm{(B)}$

See Also

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