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"

(→‎Solution 2)
Line 1: Line 1:
 
{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #17]] and [[2002 AMC 10B Problems|2002 AMC 10B #21]]}}
 
{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #17]] and [[2002 AMC 10B Problems|2002 AMC 10B #21]]}}
 
== 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 third as fast 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}

Revision as of 14:36, 1 May 2021

The following problem is from both the 2002 AMC 12B #17 and 2002 AMC 10B #21, so both problems redirect to this page.

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 1

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 time 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 $\frac{3}{4}\times\frac{x}{y}$, so the first one to finish is $\boxed{\mathrm{(B)}\ \text{Beth}}$.


Solution 2

WLOG, we can set values of their lawns' areas and their owners' speeds. Let the area of Andy's lawn be $6$ units, Beth's lawn be $3$ units, and Carlos's lawn be $2$ units. Let Carlos's mowing area per hour (honestly the time you set won't matter) be $1$, let Beth's mowing area per hour be $2$, and let Andy's mowing area per hour be $3$. Now, we can easily calculate their time by dividing their lawns' areas by their respective owners' speeds. Andy's time is $\frac{6}{3} = 2$ hours, Beth's time is $\frac{3}{2} = 1.5$ hours, and Carlos's time is $\frac{6}{3} = 2$ hours. Our answer is clearly $\boxed{\mathrm{(B)}\ \text{Beth}}$.


~Solution by virjoy2001

See also

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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
2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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 12 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=2002_AMC_12B_Problems/Problem_17&oldid=153042"