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

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{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #3]] and [[2006 AMC 10A Problems/Problem 3|2008 AMC 10A #3]]}}
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{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #3]] and [[2006 AMC 10A Problems/Problem 3|2006 AMC 10A #3]]}}
 
== Problem ==
 
== Problem ==
 
The ratio of Mary's age to Alice's age is <math>3:5</math>. Alice is <math>30</math> years old. How old is Mary?
 
The ratio of Mary's age to Alice's age is <math>3:5</math>. Alice is <math>30</math> years old. How old is Mary?

Revision as of 16:11, 1 June 2009

The following problem is from both the 2006 AMC 12A #3 and 2006 AMC 10A #3, so both problems redirect to this page.

Problem

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

$\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18\qquad\mathrm{(C)}\ 20\qquad\mathrm{(D)}\ 24\qquad\mathrm{(E)}\  50$

Solution

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

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
2006 AMC 10A (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 10 Problems and Solutions