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?

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

Solution 1

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

Solution 2

We can see this is a combined ratio of $8$, $(5+3)$. We can equalize by doing $30\div5=6$, and $6\cdot3=18$. 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.

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

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