Difference between revisions of "2005 AMC 10A Problems/Problem 7"
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==Problem== | ==Problem== | ||
− | + | 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> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } | + | <math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 4 </math> |
==Solution== | ==Solution== | ||
− | Let <math>m</math> be the distance in miles that | + | Let <math>m</math> be the distance in miles that Mike rode. |
− | Since | + | 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, | Since their combined distance was <math>13</math> miles, |
Latest revision as of 13:03, 31 May 2021
Contents
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,
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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