Difference between revisions of "2000 AMC 8 Problems/Problem 1"
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==Solution 1== | ==Solution 1== | ||
| − | If Brianna is half as old as Aunt Anna, then | + | If Brianna is half as old as Aunt Anna, then Brianna is <math>\frac{42}{2}</math> years old, or <math>21</math> years old. |
| − | If Caitlin is <math>5</math> years younger than | + | If Caitlin is <math>5</math> years younger than Brianna, she is <math>21-5</math> years old, or <math>16</math>. |
So, the answer is <math>\boxed{B}</math> | So, the answer is <math>\boxed{B}</math> | ||
Latest revision as of 03:20, 13 April 2024
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 1
If Brianna is half as old as Aunt Anna, then Brianna is 
years old, or 
years old.
If Caitlin is years younger than Brianna, she is 
years old, or 
.
So, the answer is 
Solution 2
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. 
