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

(Solution)
(Solution)
Line 7: Line 7:
 
Let <math>m</math> be the distance in miles that Mike rode.  
 
Let <math>m</math> 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 <math>2\cdot\frac{4}{5}\cdot m = \frac{8}{5}m</math> miles.
 
  
Since their combined distance was <math>13</math> miles,
 
  
 
<math> \frac{8}{5}m + m = 13 </math>
 
<math> \frac{8}{5}m + m = 13 </math>

Revision as of 20:13, 12 November 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.


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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS