Difference between revisions of "2000 AMC 8 Problems/Problem 1"
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So, the answer is <math>\boxed{B}</math> | So, the answer is <math>\boxed{B}</math> | ||
| + | ==Solution== | ||
| + | Since Brianna is half of Aunt Anna's age this means that Brianna is <math>21</math> years old. | ||
| + | Now we just find Caitlin's age by doing <math>21-5</math>. This makes <math>16</math> or <math>\boxed{B}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 19:49, 8 February 2016
Contents
Problem
Aunt Anna is 
years old. Caitlin is 
years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin?
Solution
If Brianna is half as old as Aunt Anna, then Briana is 
years old, or 
years old.
If Caitlin is years younger than Briana, she is 
years old, or 
.
So, the answer is 
Solution
Since Brianna is half of Aunt Anna's age this means that Brianna is years old.
Now we just find Caitlin's age by doing . This makes or
See Also
| 2000 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
