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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<math>\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8</math> | <math>\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8</math> | ||
| − | ==Solution== | + | ==Solutions== |
| − | <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>. | + | |
| + | ==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{\text{(C)}\ 6}</math>. | ||
| + | |||
| + | ==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>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2002|ab=A|num-b=1|num-a=3}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 22:30, 1 January 2019
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 
.
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. 
