Difference between revisions of "2005 AMC 10A Problems/Problem 7"
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<math> \frac{8}{5}m + m = 13 </math> | <math> \frac{8}{5}m + m = 13 </math> | ||
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{{AMC10 box|year=2005|ab=A|num-b=6|num-a=8}} | {{AMC10 box|year=2005|ab=A|num-b=6|num-a=8}} |
Revision as of 20:13, 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.
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,
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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