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Difference between revisions of "2013 AIME I Problems/Problem 1"

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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>
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== Solution Two ==
 
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The following solution is troll:
 
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Tom trips and dies, therefore, because the decomposition of his body is similar to the fractional decomposition of the formula <math>\frac{.5}{r} + \frac{8}{5r} + \frac{30}{10r} = 255</math>, we can easily express it in terms of e^i<math>\pi</math>theta, through Euler's formula. i is <math>\sqrt-1</math>, but we were taught in Algebra I class in 8th grade not to worry about the <math>\sqrt  </math> of negative numbers, therefore the answer is <math>\boxed{000}</math>
 
  
 
== 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}}

Revision as of 22:34, 24 March 2013

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 One

Let $r$ represent the rate Tom swims in miles per minute. Then we have

$\frac{.5}{r} + \frac{8}{5r} + \frac{30}{10r} = 255$

Solving for $r$, we find $r = 1/50$, so the time Tom spends biking is $\frac{30}{(10)(1/50)} = \boxed{150}$

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
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