Difference between revisions of "2005 AMC 10A Problems/Problem 7"
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| − | Let <math>m</math> be the distance in miles that | + | Let <math>m</math> be the distance in miles that Prthib rode. |
| − | Since | + | 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 their combined distance was <math>13</math> miles, | Since their combined distance was <math>13</math> miles, | ||
Revision as of 12:45, 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 Prthib 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
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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 | ||
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