Difference between revisions of "2002 AMC 10B Problems/Problem 21"

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== Problem ==
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== Problem 21 ==
  
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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?
  
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?
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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>
 
 
'''(A) Andy   (B) Beth   (C) Carlos   (D) Andy and Carlos tie for first.   (E) All three tie.'''
 
 
 
  
 
== Solution ==
 
== Solution ==
  
let ''l'' = area of Beth's lawn
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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>
 
 
''b'' = how fast Beth's lawn mower cuts
 
 
 
If Beth's lawn's area is ''l'' and her lawn mower's speed is ''b'', it will take her ''l''/''b'' time to finish.
 
 
 
If Andy's lawn hs twice as much area as Beth's lawn, Andy's lawn's area would be 2''l''.
 
 
 
If Andy's lawn has three times as much area as Carlos' lawn, Carlos' lawn's area would be 2''l''/3.
 
 
 
If Carlos' lawn mower cuts half as fast as Beth's mower, Carlos' mower's speed is ''b''/2.
 
 
 
If Carlos' lawn mower cuts one third as fast as Andy's mower, Andy's mower's speed is 3''b''/2.
 
 
 
Since Andy's lawn's area is 2''l'' and his lawn mower's speed is 3''b''/2, it will take him 2''l''/(3''b''/2) time to finish, or 4''l''/3''b''.
 
 
 
Since Carlos' lawn's area is 2''l''/3 and his lawn mower's speed is'' b''/2, it will take him (2''l''/3)/(''b''/2) time to finish, or 4''l''/3''b''.
 
  
''l''/''b'' < 4''l''/3''b'' = 4''l''/3''b''
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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>
  
Beth's time < Andy's time = Carlos' time
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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>
  
So, Beth finishes first. The answer is '''(B)'''
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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>

Revision as of 15:40, 4 June 2011

Problem 21

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?

$\mathrm{(A) \ } \text{Andy}\qquad \mathrm{(B) \ } \text{Beth}\qquad \mathrm{(C) \ } \text{Carlos}\qquad \mathrm{(D) \ } \text{Andy and Carlos tie for first.}\qquad \mathrm{(E) \ } \text{All three tie.}$

Solution

If we let $a$ be the area of Beth's lawn, then the area of Andy's lawn is $2a$ and the area of Carlos's lawn is $\frac{2}{3}a.$

If we let $s$ be the speed of Beth's mower, then the speed of Carlos's mower is $\frac{1}{2}s$ and the speed of Andy's mower is $\frac{3}{2}s.$

Now it follows that the time it will take to mow the lawn for Andy is $2a / \frac{3}{2}s = \frac{4}{3} \frac{a}{s}.$ The time it will take for Beth is $\frac{a}{s}$. For Carlos, it will be $\frac{2}{3}a / \frac{1}{2}s = \frac{4}{3} \frac{a}{s}.$

Since $\frac{a}{s} < \frac{4}{3} \frac{a}{s} = \frac{4}{3} \frac{a}{s}, \text{Beth} < \text{Andy} = \text{Carlos},$ and the person who finishes first is $\boxed{\mathrm{(B) \ } \text{Beth}}.$

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