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Difference between revisions of "2002 AMC 12B Problems/Problem 17"
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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>.
 
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>\frac{3}{4}\times\frac{x}{y}</math>, so the first one to finish is <math>\boxed{\mathrm{(B)}\ \text{Beth}}</math>.
+
Each person's lawn is cut at a time 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>\frac{3}{4}\times\frac{x}{y}</math>, so the first one to finish is <math>\boxed{\mathrm{(B)}\ \text{Beth}}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 14:55, 4 June 2020

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

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 $\boxed{\mathrm{(B)}\ \text{Beth}}$.

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

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