2004 AMC 8 Problem 13
Attempt Problem 13 of the 2004 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
13.
Amy, Bill and Celine are friends with different ages. Exactly one of the following statements is true.
Bill is the oldest.
Amy is not the oldest.
Celine is not the youngest.
Rank the friends from the oldest to youngest.
Bill, Amy, Celine
Amy, Bill, Celine
Celine, Amy, Bill
Celine, Bill, Amy
Amy, Celine, Bill
Answer: E
Small hint:
If Bill were oldest, check how many statements would be true.
Big hint:
After ruling out the oldest person, use the fact that exactly one statement is true.
Solution:
If Bill were oldest, then statements and would both be true, so Bill is not oldest.
If Celine were oldest, then statements and would both be true, so Celine is not oldest.
Therefore Amy is oldest. Statements and are false, so statement must be the single true statement. Thus Celine is not youngest, leaving Bill youngest.
The order is Amy, Celine, Bill.
Thus, E is the correct answer.
Problem 13 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
- 1988 AMC 8
- 1989 AMC 8
- 1990 AMC 8
- 1991 AMC 8
- 1992 AMC 8
- 1993 AMC 8
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