Difference between revisions of "2009 AMC 10A Problems/Problem 4"
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Since <math>d=rt</math>, Eric takes <math>\frac{\frac{1}{4}}{2}=\frac{1}{8}</math> hours for the swim. Then, he takes <math>\frac{3}{6}=\frac{1}{2}</math> hours for the run. So he needs to take <math>2-\frac{5}{8}=\frac{11}{8}</math> hours for the <math>15</math> mile run. This is <math>\frac{15}{\frac{11}{8}}=\frac{120}{11} \frac{\text{miles}}{\text{hour}}</math> | Since <math>d=rt</math>, Eric takes <math>\frac{\frac{1}{4}}{2}=\frac{1}{8}</math> hours for the swim. Then, he takes <math>\frac{3}{6}=\frac{1}{2}</math> hours for the run. So he needs to take <math>2-\frac{5}{8}=\frac{11}{8}</math> hours for the <math>15</math> mile run. This is <math>\frac{15}{\frac{11}{8}}=\frac{120}{11} \frac{\text{miles}}{\text{hour}}</math> | ||
Revision as of 14:11, 29 November 2020
Problem
Eric plans to compete in a triathlon. He can average 
miles per hour in the 
-mile swim and 
miles per hour in the 
-mile run. His goal is to finish the triathlon in hours. To accomplish his goal what must his average speed in miles per hour, be for the 
-mile bicycle ride?
Solution
\renewcommand{\baselinestretch}{2.0}
Since 
, Eric takes 
hours for the swim. Then, he takes 
hours for the run. So he needs to take 
hours for the mile run. This is 
See also
| 2009 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
