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> | ||
| − | <math> \frac{13}{5}m = 13 </math> | + | <math> \frac{13}{5}m = 13 </math> |
| + | <math> m = 5 \Longrightarrow \mathrm{(B)} </math> | ||
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==See also== | ==See also== | ||
{{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 21:41, 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
