Difference between revisions of "2000 AMC 8 Problems/Problem 1"
m (→Solution 1) |
(→Solution 2) |
||
| Line 13: | Line 13: | ||
==Solution 2== | ==Solution 2== | ||
Since Brianna is half of Aunt Anna's age this means that Brianna is <math>21</math> years old. | 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> | + | Now we just find Caitlin's age by doing <math>21-5</math>. This makes <math>17</math> or <math>\boxed{B}</math> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2000|before=First<br />Question|num-a=2}} | {{AMC8 box|year=2000|before=First<br />Question|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 02:17, 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. 
