# Difference between revisions of "2005 AMC 10A Problems/Problem 7"

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<math> \frac{13}{5}m = 13 </math> | <math> \frac{13}{5}m = 13 </math> | ||

− | <math> m = 5 \ | + | <math> m = 5 \Longrightarrow \mathrm{(B)} </math> |

==See Also== | ==See Also== | ||

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*[[2005 AMC 10A Problems/Problem 8|Next Problem]] | *[[2005 AMC 10A Problems/Problem 8|Next Problem]] | ||

+ | [[Category:Introductory Algebra Problems]] |

## Revision as of 10:41, 2 August 2006

## 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,