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

(Redirected page to 2002 AMC 12B Problems/Problem 17)
 
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== Problem ==
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#REDIRECT[[2002 AMC 12B Problems/Problem 17]]
 
 
 
 
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?
 
 
 
'''(A) Andy  (B) Beth  (C) Carlos  (D) Andy and Carlos tie for first.  (E) All three tie.'''
 
 
 
 
 
== Solution ==
 
 
 
let ''l''=area of Beth's lawn
 
'
 
'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''
 
 
 
Beth's time < Andy's time = Carlos' time
 
 
 
So, Beth finishes first. The answer is '''(B)'''
 

Latest revision as of 15:23, 29 July 2011

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