2024 AMC 12A Problems/Problem 10
- The following problem is from both the 2024 AMC 12A #10 and 2024 AMC 10A #13, so both problems redirect to this page.
Problem
Aubrey raced his younger brother Blair. Aubrey runs at a faster constant speed than Blair, so Blair started the race 
feet ahead of Aubrey. Aubrey caught up to Blair after 
seconds, finishing the race 
feet ahead of Blair and 
seconds earlier than Blair. How far did Aubrey run, in feet?
Solution
Assume the race starts at time 
, when Aubrey trails his younger brother Blair by feet. By time 
seconds, Aubrey has "made up" those feet, so his running speed is exactly 
feet per second faster than Blair. Therefore, if Aubrey finished the race feet ahead of Blair, he must have done so 
seconds after he caught up. This means Aubrey finished the race at time 
seconds. Suppose Aubrey ran 
feet. Then Blair finished the race at time 
seconds and ran 
feet. Aubrey was running at a speed feet per second faster throughout, and therefore ![\[\frac{d}{26} =\frac{d - 40}{31} + 5 =\frac{d + 115}{31}\]](http://latex.artofproblemsolving.com/5/9/7/597ef0f06747aa3b6407aba3845a4bd604146cee.png)
hence 
, so 
and 
.
See also
| 2024 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 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 12 Problems and Solutions | |
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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. 
