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

## 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 byProblem 19 Followed byProblem 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.

Invalid username
Login to AoPS