2006 AMC 12A Problems/Problem 3

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?

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

Solution 1

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

Solution 2

We can see this is a combined ratio of $8$ , $(5+3)$ . We can equalize by doing $30\div5=6$ , and $6\cdot3=\boxed{\textbf{(A) }18}$ . With the common ratio of $8$ and difference ratio of $6$ , we see $6\cdot8=30+18$ . Therefore, we can see our answer is correct.

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AMC 12 Problems and Solutions

2006 AMC 10A (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2006 AMC 12A Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

See also