Difference between revisions of "2005 AMC 10A Problems/Problem 7"
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Josh and Mike live <math>13</math> 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? | Josh and Mike live <math>13</math> 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? | ||
| − | <math> \ | + | <math> \textbf{(A) } 4\qquad \textbf{(B) } 5\qquad \textbf{(C) } 6\qquad \textbf{(D) } 7\qquad \textbf{(E) } 8 </math> |
==Solution== | ==Solution== | ||
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> | ||
| Line 13: | Line 15: | ||
<math> \frac{13}{5}m = 13 </math> | <math> \frac{13}{5}m = 13 </math> | ||
| − | <math> m = | + | <math> m = \boxed{\textbf{(B) }5} </math> |
| − | + | ||
==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}} | ||
| − | |||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 16:28, 10 August 2023
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
