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Difference between revisions of "2006 AMC 12A Problems/Problem 7"

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== See also ==
 
== See also ==
 
* [[2006 AMC 12A Problems]]
 
* [[2006 AMC 12A Problems]]
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*[[2006 AMC 12A Problems/Problem 6|Previous Problem]]
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*[[2006 AMC 12A Problems/Problem 8|Next Problem]]
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[[Category:Introductory Algebra Problems]]

Revision as of 18:50, 5 November 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 $23.2$ years. How old will Mary be on her next birthday?

$\mathrm{(A) \ } 7\qquad \mathrm{(B) \ } 8\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 11$

Solution

Let $m$ be Mary's age, let $s$ be Sally's age, and let $d$ be Danielle's age. We have $s=.6d$, and $m=1.2s=1.2(.6d)=.72d$. The sum of their ages is $m+s+d=.72d+.6d+d=2.32d$. Therefore, $2.32d=23.2$, and $d=10$. Then $m=.72(10)=7.2$. Mary will be $8$ on her next birthday. The answer is B.

See also

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