Difference between revisions of "2013 AIME I Problems/Problem 1"
m (→Solution) |
m (→Solution) |
||
| Line 8: | Line 8: | ||
<math>\frac{.5}{r} + \frac{8}{5r} + \frac{30}{10r} = 255</math> | <math>\frac{.5}{r} + \frac{8}{5r} + \frac{30}{10r} = 255</math> | ||
| − | Solving for <math>r</math>, we find <math>r = 1/50</math>, so the time Tom spends biking is <math>\frac{30}{(10)(1/50)} = \boxed{150}</math> | + | Solving for <math>r</math>, we find <math>r = 1/50</math>, so the time Tom spends biking is <math>\frac{30}{(10)(1/50)} = \boxed{150}</math> minutes. |
== See also == | == See also == | ||
{{AIME box|year=2013|n=I|before=First Problem|num-a=2}} | {{AIME box|year=2013|n=I|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 09:05, 24 May 2018
Problem 1
The AIME Triathlon consists of a half-mile swim, a 30-mile bicycle ride, and an eight-mile run. Tom swims, bicycles, and runs at constant rates. He runs fives times as fast as he swims, and he bicycles twice as fast as he runs. Tom completes the AIME Triathlon in four and a quarter hours. How many minutes does he spend bicycling?
Solution
Let 
represent the rate Tom swims in miles per minute. Then we have
Solving for , we find 
, so the time Tom spends biking is 
minutes.
See also
| 2013 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
