Difference between revisions of "2006 AMC 12A Problems/Problem 3"

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== See also ==
 
== See also ==
 
* [[2006 AMC 12A Problems]]
 
* [[2006 AMC 12A Problems]]
*[[2006 AMC 12A Problems/Problem 2|Previous Problem]]
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*[[2006 AMC 12A Problems/Problem 4|Next Problem]]
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{{AMC12 box|year=2006|ab=A|num-b=2|num-a=4}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 19:28, 2 February 2007

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

2006 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions