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2006 AMC 12A Problems/Problem 3: Difference between revisions

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Solution
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== See also ==
== See also ==
* [[2006 AMC 12A Problems]]
* [[2006 AMC 12A Problems]]
*[[2006 AMC 12A Problems/Problem 2|Previous Problem]]
*[[2006 AMC 12A Problems/Problem 4|Next Problem]]
[[Category:Introductory Algebra Problems]]

Revision as of 17:44, 5 November 2006

Problem

The ratio of Mary's age to Alice's age is $3:5$. Alice is $30$ years old. How old is Mary?

$\mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 18\qquad \mathrm{(C) \ } 20\qquad \mathrm{(D) \ } 24$

$\mathrm{(E) \ } 50$

Solution

Let $m$ be Mary's age. Then $\frac{m}{30}=\frac{3}{5}$. Solving for $m$, we obtain $m=18$. The answer is B.

See also

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