Difference between revisions of "2009 AMC 10A Problems/Problem 4"
VelaDabant (talk | contribs) (New page: ==Problem== Eric plans to compete in a triathlon. He can average <math>2</math> miles per hour in the <math>\frac{1}{4}</math>-mile swim and <math>6</math> miles per hour in the <math>3</m...) |
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==Solution== | ==Solution== | ||
+ | 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> | ||
+ | |||
<math>\longrightarrow \fbox{A}</math> | <math>\longrightarrow \fbox{A}</math> | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2009|ab=A|num-b=3|num-a=5}} | {{AMC10 box|year=2009|ab=A|num-b=3|num-a=5}} |
Revision as of 09:27, 9 February 2010
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
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 |