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Difference between revisions of "2008 AMC 12A Problems/Problem 2"

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<math>\mathrm{(A)}\ \frac{6}{7}\qquad\mathrm{(B)}\ \frac{7}{6}\qquad\mathrm{(C)}\ \frac{5}{3}\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ \frac{7}{2}</math>
 
<math>\mathrm{(A)}\ \frac{6}{7}\qquad\mathrm{(B)}\ \frac{7}{6}\qquad\mathrm{(C)}\ \frac{5}{3}\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ \frac{7}{2}</math>
  
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==Solution==
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==Solution 1==
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===Solution 1===
 
  
 
Here's a cheapshot:  
 
Here's a cheapshot:  
 
Obviously, <math>\frac{1}{2}+\frac{2}{3}</math> is greater than <math>1</math>. Therefore, its reciprocal is less than <math>1</math>, and the answer must be <math>\boxed{\frac{6}{7}}</math>.
 
Obviously, <math>\frac{1}{2}+\frac{2}{3}</math> is greater than <math>1</math>. Therefore, its reciprocal is less than <math>1</math>, and the answer must be <math>\boxed{\frac{6}{7}}</math>.
  
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===Solution 2===
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==Solution 2==
  
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<math>\left(\frac{1}{2}+\frac{2}{3}\right)^{-1}=\left(\frac{3}{6}+\frac{4}{6}\right)^{-1}=\left(\frac{7}{6}\right)^{-1}=\frac{6}{7}\Rightarrow A</math>.
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<math>\left(\frac{1}{2}+\frac{2}{3}\right)^{-1}=\left(\frac{3}{6}+\frac{4}{6}\right)^{-1}=\left(\frac{7}{6}\right)^{-1}=\boxed{\mathrm{(A)}\ \frac{6}{7}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2008|ab=A|num-b=1|num-a=3}}
 
{{AMC12 box|year=2008|ab=A|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:22, 7 September 2021

Problem

What is the reciprocal of $\frac{1}{2}+\frac{2}{3}$?

$\mathrm{(A)}\ \frac{6}{7}\qquad\mathrm{(B)}\ \frac{7}{6}\qquad\mathrm{(C)}\ \frac{5}{3}\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ \frac{7}{2}$

Solution 1

Here's a cheapshot: Obviously, $\frac{1}{2}+\frac{2}{3}$ is greater than $1$. Therefore, its reciprocal is less than $1$, and the answer must be $\boxed{\frac{6}{7}}$.

Solution 2

$\left(\frac{1}{2}+\frac{2}{3}\right)^{-1}=\left(\frac{3}{6}+\frac{4}{6}\right)^{-1}=\left(\frac{7}{6}\right)^{-1}=\boxed{\mathrm{(A)}\ \frac{6}{7}}$.

See Also

2008 AMC 12A (Problems • Answer Key • Resources)
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Problem 3
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All AMC 12 Problems and Solutions

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