2002 AMC 10A Problems/Problem 2

Problem

Given that a, b, and c are non-zero real numbers, define $(a, b, c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$, find $(2, 12, 9)$.

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$

Solutions

Solution 1

$(2, 12, 9)=\frac{2}{12}+\frac{12}{9}+\frac{9}{2}=\frac{1}{6}+\frac{4}{3}+\frac{9}{2}=\frac{1}{6}+\frac{8}{6}+\frac{27}{6}=\frac{36}{6}=6$. Our answer is then $\boxed{\textbf{(C) }6}$.

Solution 2

Without computing the answer exactly, we see that $2/12=\text{a little}$, $12/9=\text{more than }1$, and $9/2=4.5$. The sum is $4.5 + (\text{more than }1) + (\text{a little}) = (\text{more than }5.5) + (\text{a little})$, and as all the options are integers, the correct one is $\boxed{\textbf{(C) }6}$.

Video Solution by Daily Dose of Math

https://youtu.be/I1LoajlCAgg

~Thesmartgreekmathdude

See Also

2002 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 10 Problems and Solutions

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