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| − | == Problem 21 ==
| + | #REDIRECT[[2002 AMC 12B Problems/Problem 17]] |
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| − | Andy's lawn has twice as much area as Beth's lawn and three times as much 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?
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| − | <math> \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.} </math>
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| − | == Solution ==
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| − | If we let <math>a</math> be the area of Beth's lawn, then the area of Andy's lawn is <math>2a</math> and the area of Carlos's lawn is <math>\frac{2}{3}a.</math>
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| − | If we let <math>s</math> be the speed of Beth's mower, then the speed of Carlos's mower is <math>\frac{1}{2}s</math> and the speed of Andy's mower is <math>\frac{3}{2}s.</math>
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| − | Now it follows that the time it will take to mow the lawn for Andy is <math>2a / \frac{3}{2}s = \frac{4}{3} \frac{a}{s}.</math> The time it will take for Beth is <math>\frac{a}{s}</math>. For Carlos, it will be <math>\frac{2}{3}a / \frac{1}{2}s = \frac{4}{3} \frac{a}{s}.</math>
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| − | Since <math>\frac{a}{s} < \frac{4}{3} \frac{a}{s} = \frac{4}{3} \frac{a}{s}, \text{Beth} < \text{Andy} = \text{Carlos},</math> and the person who finishes first is <math>\boxed{\mathrm{(B) \ } \text{Beth}}.</math>
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