Difference between revisions of "2005 AMC 10A Problems/Problem 7"
(→Problem) |
(→Solution) |
||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| − | Let <math>m</math> be the distance in miles that | + | Let <math>m</math> be the distance in miles that Parthib rode. |
Since Wahida rode for twice the length of time as Parthib and at four-fifths of Parthib's rate, he rode <math>2\cdot\frac{4}{5}\cdot m = \frac{8}{5}m</math> miles. | Since Wahida rode for twice the length of time as Parthib and at four-fifths of Parthib's rate, he rode <math>2\cdot\frac{4}{5}\cdot m = \frac{8}{5}m</math> miles. | ||
Revision as of 12:48, 17 February 2021
Contents
Problem
Wahida and Parthib live 
miles apart. Yesterday Wahida started to ride his bicycle toward Parthib's house. A little later Parthib started to ride his bicycle toward Wahida's house. When they met, Wahida had ridden for twice the length of time as Parthib and at four-fifths of Parthib's rate. How many miles had Parthib ridden when they met?
Solution
Let 
be the distance in miles that Parthib rode.
Since Wahida rode for twice the length of time as Parthib and at four-fifths of Parthib's rate, he rode 
miles.
Since their combined distance was miles,
Video Solution
-Check this amazing solution: https://youtu.be/WIR8yPLET9Y
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
