Difference between revisions of "2009 AMC 10A Problems/Problem 4"
(→Solution) |
(→Solution) |
||
(3 intermediate revisions by one other user not shown) | |||
Line 15: | Line 15: | ||
==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> | + | 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. |
− | hours for the run. | ||
− | <math>\longrightarrow \fbox{A}</math> | + | So he needs to take <math>2-\frac{5}{8}=\frac{11}{8}</math> hours for the <math>15</math> mile bike. This is <math>\frac{15}{\frac{11}{8}}=\frac{120}{11} \frac{\text{miles}}{\text{hour}}</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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:30, 11 September 2022
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 bike. 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.