Difference between revisions of "2016 AMC 8 Problems/Problem 16"

(Solution 1)
 
(2 intermediate revisions by one other user not shown)
Line 6: Line 6:
  
 
==Solutions==
 
==Solutions==
Solution 1
+
===Solution 1===
 
Each lap Bonnie runs, Annie runs another quarter lap, so Bonnie will run four laps before she is overtaken.  This means that Annie and Bonnie are equal so that Annie needs to run another lap to overtake Bonnie. That means Annie will have run <math>\boxed{\textbf{(D)}\ 5 }</math> laps.
 
Each lap Bonnie runs, Annie runs another quarter lap, so Bonnie will run four laps before she is overtaken.  This means that Annie and Bonnie are equal so that Annie needs to run another lap to overtake Bonnie. That means Annie will have run <math>\boxed{\textbf{(D)}\ 5 }</math> laps.
  
==Video Solution 1==
+
== Video Solution by Pi Academy ==
 +
 
 +
https://youtu.be/Sn1QLgCJUAQ?si=UebITjxagXo7KoDI
 +
 
 +
 
 +
==Video Solution 2==
 
https://youtu.be/lRbxzdBZpIY
 
https://youtu.be/lRbxzdBZpIY
  

Latest revision as of 18:09, 23 November 2024

Problem

Annie and Bonnie are running laps around a $400$-meter oval track. They started together, but Annie has pulled ahead, because she runs $25\%$ faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?

$\textbf{(A) }1\dfrac{1}{4}\qquad\textbf{(B) }3\dfrac{1}{3}\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }25$

Solutions

Solution 1

Each lap Bonnie runs, Annie runs another quarter lap, so Bonnie will run four laps before she is overtaken. This means that Annie and Bonnie are equal so that Annie needs to run another lap to overtake Bonnie. That means Annie will have run $\boxed{\textbf{(D)}\ 5 }$ laps.

Video Solution by Pi Academy

https://youtu.be/Sn1QLgCJUAQ?si=UebITjxagXo7KoDI


Video Solution 2

https://youtu.be/lRbxzdBZpIY

~savannahsolver

2016 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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