2004 AMC 8 Problems/Problem 13

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.

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

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

Solution 2

Bashing through/ eliminating the options

The first statement tells us that bill is the oldest. That is only what we need to know to solve now. In the answer choices, you can easily see that the first one is the only option where Bill is the oldest. Thus, the answer is $\boxed{\textbf{(E)}\ \text{Bill, Celine, Bill}}$ .

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.

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2004 AMC 8 Problems/Problem 13

Contents

Problem

Solution

Solution 2

See Also