2002 AMC 12B Problems/Problem 17

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 $\Rightarrow \mathrm{(B)}$ .

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2002 AMC 12B Problems/Problem 17

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