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 "2002 AMC 12B Problems/Problem 17"

(→‎Problem)
(→‎Problem)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
Andy’s lawn has twice as much area as Beth’s lawn and three times as much area as Carlos’ lawn. Carlos’ lawn mower cuts half as fast as Beth’s mower and one thid as afst as Andy’s mower. If they all start to mow their lawns at the same time, who will finish first?
+
Andy’s lawn has twice as much area as Beth’s lawn and three times as much area as Carlos’ lawn. Carlos’ lawn mower cuts half as fast as Beth’s mower and one third as fast as Andy’s mower. If they all start to mow their lawns at the same time, who will finish first?
  
 
<math>\mathrm{(A)}\ \text{Andy}
 
<math>\mathrm{(A)}\ \text{Andy}
Line 7: Line 7:
 
<math>\qquad\mathrm{(D)}\ \text{Andy\ and \ Carlos\ tie\ for\ first.}
 
<math>\qquad\mathrm{(D)}\ \text{Andy\ and \ Carlos\ tie\ for\ first.}
 
\qquad\mathrm{(E)}\ \text{All\ three\ tie.}</math>
 
\qquad\mathrm{(E)}\ \text{All\ three\ tie.}</math>
 +
 +
== Solution ==
 +
We say Andy's lawn has an area of <math>x</math>. Beth's lawn thus has an area of <math>\frac{x}2</math>, and Carlos's lawn has an area of <math>\frac{x}3</math>.
 +
 +
We say Andy's lawn mower cuts at a speed of <math>y</math>. Carlos's cuts at a speed of <math>\frac{y}3</math>, and Beth's cuts at a speed <math>\frac{2y}3</math>.
 +
 +
Each person's lawn is cut at a speed of <math>\frac{\text{area}}{\text{rate}}</math>, so Andy's is cut in <math>\frac{x}y</math> time, as is Carlos's. Beth's is cut in <math>\frac34*\frac{x}y</math>, so Beth finishes first <math>\Rightarrow \mathrm{(B)}</math>.

Revision as of 15:52, 17 February 2008

Problem

Andy’s lawn has twice as much area as Beth’s lawn and three times as much area as Carlos’ lawn. Carlos’ lawn mower cuts half as fast as Beth’s mower and one third as fast as Andy’s mower. If they all start to mow their lawns at the same time, who will finish first?

$\mathrm{(A)}\ \text{Andy} \qquad\mathrm{(B)}\ \text{Beth} \qquad\mathrm{(C)}\ \text{Carlos}$ $\qquad\mathrm{(D)}\ \text{Andy\ and \ Carlos\ tie\ for\ first.} \qquad\mathrm{(E)}\ \text{All\ three\ tie.}$

Solution

We say Andy's lawn has an area of $x$. Beth's lawn thus has an area of $\frac{x}2$, and Carlos's lawn has an area of $\frac{x}3$.

We say Andy's lawn mower cuts at a speed of $y$. Carlos's cuts at a speed of $\frac{y}3$, and Beth's cuts at a speed $\frac{2y}3$.

Each person's lawn is cut at a speed of $\frac{\text{area}}{\text{rate}}$, so Andy's is cut in $\frac{x}y$ time, as is Carlos's. Beth's is cut in $\frac34*\frac{x}y$, so Beth finishes first $\Rightarrow \mathrm{(B)}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_17&oldid=23408"