Difference between revisions of "2005 AMC 10A Problems/Problem 7"
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| − | Let <math>m</math> be the distance in miles that | + | Let <math>m</math> be the distance in miles that Mike rode. |
| − | Since | + | 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, | Since their combined distance was <math>13</math> miles, | ||
Revision as of 12:03, 31 May 2021
Contents
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.
Since Josh rode for twice the length of time as Mike and at four-fifths of Mike's rate, he rode 
miles.
Since their combined distance was miles,
Video Solution
-Check this amazing solution: https://youtu.be/WIR8yPLET9Y
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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