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

(New page: ==Problem== Given that a, b, and c are non-zero real numbers, define <math>(a, b, c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}</math>, find <math>(2, 12, 9)</math>. <math>\text{(A)}\ 4 \q...)
 
 
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Given that a, b, and c are non-zero real numbers, define <math>(a, b, c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}</math>, find <math>(2, 12, 9)</math>.
 
Given that a, b, and c are non-zero real numbers, define <math>(a, b, c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}</math>, find <math>(2, 12, 9)</math>.
  
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<math>\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8</math>
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<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8</math>
  
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==Solution==
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==Solutions==
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<math>(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}=\boxed{6}</math>. Our answer is then <math>\text{(C)}\ 6 \qquad</math>.
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==Solution 1==
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<math>(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</math>. Our answer is then <math>\boxed{\textbf{(C) }6}</math>.
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==Solution 2==
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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>\boxed{\textbf{(C) }6}</math>.
 +
 
 +
==Video Solution by Daily Dose of Math==
 +
 
 +
https://youtu.be/I1LoajlCAgg
 +
 
 +
~Thesmartgreekmathdude
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==See Also==
 +
{{AMC10 box|year=2002|ab=A|num-b=1|num-a=3}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 23:41, 18 July 2024

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 (Problems • Answer Key • Resources)
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Problem 3
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All AMC 10 Problems and Solutions

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