2004 AMC 8 Problems/Problem 13


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.

$\textbf{(A)}\ \text{Bill, Amy, Celine}\qquad \textbf{(B)}\ \text{Amy, Bill, Celine}\qquad \textbf{(C)}\ \text{Celine, Amy, Bill}\\ \textbf{(D)}\ \text{Celine, Bill, Amy} \qquad \textbf{(E)}\ \text{Amy, Celine, Bill}$


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 $\boxed{\textbf{(E)}\ \text{Amy, Celine, Bill}}$.

Solution 2

Bashing through/ eliminating the options

The first statement tells us that bill is the oldest. In the answer choices, you can see that the first answer is the only option where Bill is the oldest. Bill being the oldest is false, so we can eliminate that answer. Now we know that Bill isn't the oldest and could be the youngest or the second oldest. Don't forget, we have TWO false statements and only one true statement. The second statement where Amy is not the oldest is another false statement. So now we know Amy is the oldest, followed by Celine whom is the second oldest, and Bill, whom is the youngest.Thus, the answer is $\boxed{\textbf{(E)}\ \text{Amy, Celine, Bill}}$.



See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
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

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