Difference between revisions of "2000 AMC 8 Problems/Problem 1"
Pidigits125 (talk | contribs) (→See Also) |
|||
(13 intermediate revisions by 10 users not shown) | |||
Line 2: | Line 2: | ||
Aunt Anna is <math>42</math> years old. Caitlin is <math>5</math> years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin? | Aunt Anna is <math>42</math> years old. Caitlin is <math>5</math> years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin? | ||
− | <math> | + | <math>\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 16\qquad\mathrm{(C)}\ 17\qquad\mathrm{(D)}\ 21\qquad\mathrm{(E)}\ 37</math> |
− | \mathrm{(A)}\ 15 | ||
− | \qquad | ||
− | \mathrm{(B)}\ 16 | ||
− | \qquad | ||
− | \mathrm{(C)}\ 17 | ||
− | \qquad | ||
− | \mathrm{(D)}\ 21 | ||
− | \qquad | ||
− | \mathrm{(E)}\ 37 | ||
− | </math> | ||
− | |||
− | |||
− | If Caitlin is <math>5</math> years younger than | + | ==Solution 1== |
+ | 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 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> | ||
==See Also== | ==See Also== | ||
− | + | {{AMC8 box|year=2000|before=First<br />Question|num-a=2}} | |
− | + | {{MAA Notice}} | |
− |
Latest revision as of 02:58, 28 June 2024
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
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.