2006 AMC 12A Problems/Problem 3
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(Redirected from 2006 AMC 10A 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 
. Alice is 
years old. How old is Mary?
Solution
Let 
be Mary's age. Then 
. Solving for , we obtain 
. The answer is 
.
See also
| 2006 AMC 12A (Problems • 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 | ||
| 2006 AMC 10A (Problems • 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 | ||
