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

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<math> m = 5 \Longrightarrow \mathrm{(B)} </math>
 
<math> m = 5 \Longrightarrow \mathrm{(B)} </math>
 
   
 
   
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==See also==
*[[2005 AMC 10A Problems]]
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{{AMC10 box|year=2005|ab=A|num-b=6|num-a=8}}
  
*[[2005 AMC 10A Problems/Problem 6|Previous Problem]]
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[[Category:Introductory Number Theory Problems]]
 
 
*[[2005 AMC 10A Problems/Problem 8|Next Problem]]
 
[[Category:Introductory Algebra Problems]]
 
 
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Revision as of 14:26, 13 August 2019

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

2005 AMC 10A (ProblemsAnswer KeyResources)
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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