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 "2010 AMC 8 Problems/Problem 8"

m (→‎Solution)
(→‎Solution)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
−
Because they are both moving in the same direction, Emily is riding relative to Emerson <math>12-8=4</math> mph. Now we can look at it as if Emerson is not moving at all[on his skateboard] and Emily is riding at <math>4</math> mph. It takes her
+
Because they are both moving in the same direction, Emily is riding relative to Emerson 12-8=4 mph Now we can look at it as if Emerson is not moving at all[on his skateboard] and Emily is riding at 4 mph. It takes her
 +
1/2mile*1hour/4miles =1/8 hour to ride the 1/2 mile to reach him, and then the same amount of time to be 1/2 mile ahead of him. This totals to
  
−
<cmath>\frac12 \ \text{mile} \cdot \frac{1\ \text{hour}}{4\ \text{miles}} = \frac18\ \text{hour}</cmath>
+
2*(1/8hour)*60 min/1 hour =15 min
−
 
 
−
to ride the <math>1/2</math> mile to reach him, and then the same amount of time to be <math>1/2</math> mile ahead of him. This totals to
 
−
 
 
−
<cmath>2 \cdot \frac18 \ \text{hour} \cdot \frac{60\ \text{minutes}}{1\ \text{hour}} = \boxed{\textbf{(D)}\ 15}\ \text{minutes}</cmath>.
 
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2010|num-b=7|num-a=9}}
 
{{AMC8 box|year=2010|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:07, 15 August 2021

Problem

As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction $1/2$ mile in front of her. After she passes him, she can see him in her rear mirror until he is $1/2$ mile behind her. Emily rides at a constant rate of $12$ miles per hour, and Emerson skates at a constant rate of $8$ miles per hour. For how many minutes can Emily see Emerson?

$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16$

Solution

Because they are both moving in the same direction, Emily is riding relative to Emerson 12-8=4 mph Now we can look at it as if Emerson is not moving at all[on his skateboard] and Emily is riding at 4 mph. It takes her 1/2mile*1hour/4miles =1/8 hour to ride the 1/2 mile to reach him, and then the same amount of time to be 1/2 mile ahead of him. This totals to

2*(1/8hour)*60 min/1 hour =15 min

See Also

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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 AJHSME/AMC 8 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=2010_AMC_8_Problems/Problem_8&oldid=160335"