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Difference between revisions of "2002 AMC 10A Problems/Problem 2"

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Revision as of 11:12, 4 July 2013

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 $(2, 12, 9)$.

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

Solution

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

Alternate solution for the lazy: 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 obviously $6$.

See Also

2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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 AMC 10 Problems and Solutions

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