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.
Contents
Problem
The ratio of Mary's age to Alice's age is . Alice is years old. How old is Mary?
Solution 1
Let be Mary's age. Then . Solving for , we obtain
Solution 2
We can see this is a combined ratio of , . We can equalize by doing , and . With the common ratio of and difference ratio of , we see . Therefore, we can see our answer is correct.
See also
2006 AMC 12A (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 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.