Difference between revisions of "2004 AMC 8 Problems/Problem 13"
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II. Amy is not the oldest. | II. Amy is not the oldest. | ||
III. Celine is not the youngest. | III. Celine is not the youngest. | ||
| − | Rank the friends from the oldest to youngest | + | Rank the friends from the oldest to youngest. |
<math>\textbf{(A)}\ \text{Bill, Amy, Celine}\qquad \textbf{(B)}\ \text{Amy, Bill, Celine}\qquad \textbf{(C)}\ \text{Celine, Amy, Bill}\\ | <math>\textbf{(A)}\ \text{Bill, Amy, Celine}\qquad \textbf{(B)}\ \text{Amy, Bill, Celine}\qquad \textbf{(C)}\ \text{Celine, Amy, Bill}\\ | ||
Revision as of 18:35, 25 September 2019
Problem
Amy, Bill and Celine are friends with different ages. Exactly one of the following statements is true.
I. Bill is the oldest. II. Amy is not the oldest. III. Celine is not the youngest.
Rank the friends from the oldest to youngest.
Solution
If Bill is the oldest, then Amy is not the oldest, and both statements I and II are true, so statement I is not the true one.
If Amy is not the oldest, and we know Bill cannot be the oldest, then Celine is the oldest. This would mean she is not the youngest, and both statements II and III are true, so statement II is not the true one.
Therefore, statement III is the true statement, and both I and II are false. From this, Amy is the oldest, Celine is in the middle, and lastly Bill is the youngest. This order is 
.
See Also
| 2004 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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. 
