Difference between revisions of "2002 AMC 10A Problems/Problem 2"
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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>. | ||
| − | <math>\ | + | <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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| + | ==Solutions== | ||
| − | |||
==Solution 1== | ==Solution 1== | ||
| − | <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{\ | + | <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>. |
==Solution 2== | ==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>. |
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| + | ==Video Solution by Daily Dose of Math== | ||
| + | |||
| + | https://youtu.be/I1LoajlCAgg | ||
| + | |||
| + | ~Thesmartgreekmathdude | ||
==See Also== | ==See Also== | ||
Latest revision as of 23:41, 18 July 2024
Contents
Problem
Given that a, b, and c are non-zero real numbers, define 
, find 
.
Solutions
Solution 1
. Our answer is then 
.
Solution 2
Without computing the answer exactly, we see that 
, 
, and 
.
The sum is 
, and as all the options are integers, the correct one is .
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
