Difference between revisions of "2006 AMC 12A Problems/Problem 7"
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== Solution == | == Solution == | ||
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| + | Let <math>m</math> be Mary's age, let <math>s</math> be Sally's age, and let <math>d</math> be Danielle's age. We have <math>s=.6d</math>, and <math>m=1.2s=1.2(.6d)=.72d</math>. The sum of their ages is <math>m+s+d=.72d+.6d+d=2.32d</math>. Therefore, <math>2.32d=23.2</math>, and <math>d=10</math>. Then <math>m=.72(10)=7.2</math>. Mary will be <math>8</math> on her next birthday. The answer is B. | ||
== See also == | == See also == | ||
* [[2006 AMC 12A Problems]] | * [[2006 AMC 12A Problems]] | ||
Revision as of 14:14, 12 July 2006
Problem
Mary is $20%$ (Error compiling LaTeX. Unknown error_msg) older than Sally, and Sally is $40%$ (Error compiling LaTeX. Unknown error_msg) younger than Danielle. The sum of their ages is 
years. How old will Mary be on her next birthday?
Solution
Let 
be Mary's age, let 
be Sally's age, and let 
be Danielle's age. We have 
, and 
. The sum of their ages is 
. Therefore, 
, and 
. Then 
. Mary will be 
on her next birthday. The answer is B.
