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

Revision as of 11:26, 8 November 2021

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 $\boxed{\textbf{(C) }6}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 11: / Line 11: @@
 ==Solution 2==
 Without computing the answer exactly, we see that <math>2/12=\text{a little}</math>, <math>12/9=\text{more than }1</math>, and <math>9/2=4.5</math>.
-The sum is <math>4.5 + (\text{more than }1) + (\text{a little}) = (\text{more than }5.5) + (\text{a little})</math>, and as all the options are integers, the correct one is <math>6</math>.
+The sum is <math>4.5 + (\text{more than }1) + (\text{a little}) = (\text{more than }5.5) + (\text{a little})</math>, and as all the options are integers, the correct one is <math>\boxed{\textbf{(C) }6}</math>.
 ==See Also==

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