2005 AMC 10A Problem 7

Below is the professionally curated solution for Problem 7 of the 2005 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 10A solutions, or check the answer key.

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Concepts:distance rate and time linear equation

Difficulty rating: 1240

Josh and Mike live $13$ 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?

$4$

$5$

$6$

$7$

$8$

Solution:

Let Mike ride $m$ miles. Josh rides $\frac{4}{5}$ the rate for $2$ times the time, so Josh's distance is $\frac{8}{5}m.$ Together they cover $13,$ so $m + \frac{8}{5}m = \frac{13}{5}m = 13,$ giving $m = 5.$

Thus, the correct answer is B.

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