Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2009 AMC 10A Problems/Problem 4"

(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...)
 
(→‎Solution)
 
(11 intermediate revisions by 5 users not shown)
Line 15: Line 15:
  
 
==Solution==
 
==Solution==
−
<math>\longrightarrow \fbox{A}</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.
 +
 
 +
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}}

Latest revision as of 13:30, 11 September 2022

Problem

Eric plans to compete in a triathlon. He can average $2$ miles per hour in the $\frac{1}{4}$-mile swim and $6$ miles per hour in the $3$-mile run. His goal is to finish the triathlon in $2$ hours. To accomplish his goal what must his average speed in miles per hour, be for the $15$-mile bicycle ride?

$\mathrm{(A)}\ \frac{120}{11} \qquad \mathrm{(B)}\ 11 \qquad \mathrm{(C)}\ \frac{56}{5} \qquad \mathrm{(D)}\ \frac{45}{4} \qquad \mathrm{(E)}\ 12$

Solution

Since $d=rt$, Eric takes $\frac{\frac{1}{4}}{2}=\frac{1}{8}$ hours for the swim. Then, he takes $\frac{3}{6}=\frac{1}{2}$ hours for the run.

So he needs to take $2-\frac{5}{8}=\frac{11}{8}$ hours for the $15$ mile bike. This is $\frac{15}{\frac{11}{8}}=\frac{120}{11} \frac{\text{miles}}{\text{hour}}$$\longrightarrow \fbox{A}$

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_4&oldid=178125"