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Difference between revisions of "2012 AMC 8 Problems/Problem 20"

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<math> \textbf{(D)}\hspace{.05in}\frac{5}{19}<\frac{9}{23}<\frac{7}{21}\quad\textbf{(E)}\hspace{.05in}\frac{7}{21}<\frac{5}{19}<\frac{9}{23} </math>
 
<math> \textbf{(D)}\hspace{.05in}\frac{5}{19}<\frac{9}{23}<\frac{7}{21}\quad\textbf{(E)}\hspace{.05in}\frac{7}{21}<\frac{5}{19}<\frac{9}{23} </math>
  
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==Solution==
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==Solution 1==
 
The value of <math> \frac{7}{21} </math> is <math> \frac{1}{3} </math>. Now we give all the fractions a common denominator.
 
The value of <math> \frac{7}{21} </math> is <math> \frac{1}{3} </math>. Now we give all the fractions a common denominator.
  
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Ordering the fractions from least to greatest, we find that they are in the order listed. Therefore, our final answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>.
 
Ordering the fractions from least to greatest, we find that they are in the order listed. Therefore, our final answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>.
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 +
==Solution 2==
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Instead of finding the LCD, we can subtract each fraction from <math>1</math> to get a common numerator. Thus,
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 +
<math>1-\dfrac{5}{19}=\dfrac{14}{19}</math>
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<math>1-\dfrac{7}{21}=\dfrac{14}{21}</math>
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 +
<math>1-\dfrac{9}{23}=\dfrac{14}{23}</math>
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All three fractions have common denominator <math>14</math>. Now it is obvious the order of the fractions. <math>\dfrac{14}{19}<\dfrac{14}{21}<\dfrac{14}{23}</math>. Thereforem our answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=19|num-a=21}}
 
{{AMC8 box|year=2012|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:21, 11 November 2013

Problem

What is the correct ordering of the three numbers $\frac{5}{19}$, $\frac{7}{21}$, and $\frac{9}{23}$, in increasing order?

$\textbf{(A)}\hspace{.05in}\frac{9}{23}<\frac{7}{21}<\frac{9}{23}\quad\textbf{(B)}\hspace{.05in}\frac{5}{19}<\frac{7}{21}<\frac{9}{23}\quad\textbf{(C)}\hspace{.05in}\frac{9}{23}<\frac{5}{19}<\frac{7}{21}$

$\textbf{(D)}\hspace{.05in}\frac{5}{19}<\frac{9}{23}<\frac{7}{21}\quad\textbf{(E)}\hspace{.05in}\frac{7}{21}<\frac{5}{19}<\frac{9}{23}$

Solution 1

The value of $\frac{7}{21}$ is $\frac{1}{3}$. Now we give all the fractions a common denominator.

$\frac{5}{19} \implies \frac{345}{1311}$

$\frac{1}{3} \implies \frac{437}{1311}$

$\frac{9}{23} \implies \frac{513}{1311}$

Ordering the fractions from least to greatest, we find that they are in the order listed. Therefore, our final answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$.

Solution 2

Instead of finding the LCD, we can subtract each fraction from $1$ to get a common numerator. Thus,

$1-\dfrac{5}{19}=\dfrac{14}{19}$

$1-\dfrac{7}{21}=\dfrac{14}{21}$

$1-\dfrac{9}{23}=\dfrac{14}{23}$

All three fractions have common denominator $14$. Now it is obvious the order of the fractions. $\dfrac{14}{19}<\dfrac{14}{21}<\dfrac{14}{23}$. Thereforem our answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$.

See Also

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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