Difference between revisions of "2002 AMC 12B Problems/Problem 17"
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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>. | 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>. | ||
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| + | == See also == | ||
| + | {{AMC12 box|year=2002|ab=B|num-b=16|num-a=18}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
Revision as of 16:21, 21 February 2010
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?
Solution
We say Andy's lawn has an area of 
. Beth's lawn thus has an area of 
, and Carlos's lawn has an area of 
.
We say Andy's lawn mower cuts at a speed of 
. Carlos's cuts at a speed of 
, and Beth's cuts at a speed 
.
Each person's lawn is cut at a speed of 
, so Andy's is cut in 
time, as is Carlos's. Beth's is cut in 
, so Beth finishes first 
.
See also
| 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 | |
