2005 AMC 10A Problems/Problem 7: Difference between revisions
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Let <math>m</math> be the distance in miles that Mike rode. | Let <math>m</math> be the distance in miles that Mike rode. | ||
<math> \frac{8}{5}m + m = 13 </math> | <math> \frac{8}{5}m + m = 13 </math> | ||
Revision as of 19:13, 12 November 2019
Problem
Josh and Mike live 
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?
Solution
Let 
be the distance in miles that Mike rode.
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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