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

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Alternate solution for the lazy: 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>.
 
Alternate solution for the lazy: 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 obviously <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>6</math>.
  
 
==See Also==
 
==See Also==

Revision as of 17:07, 21 November 2018

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 $6$.

See Also

2002 AMC 10A (ProblemsAnswer KeyResources)
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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