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?

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

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

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

Change $7/21$ into $1/3$ ; $\frac{1}{3}\cdot\frac{5}{5}=\frac{5}{15}$ $\frac{5}{15}>\frac{5}{19}$ $\frac{7}{21}>\frac{5}{19}$ And $\frac{1}{3}\cdot\frac{9}{9}=\frac{9}{27}$ $\frac{9}{27}<\frac{9}{23}$ $\frac{7}{21}<\frac{9}{23}$ Therefore, our answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$ .

Solution 3

We know that $\frac{5}{19}$ is $\frac{14}{19}$ away from 0, $\frac{7}{21}$ is $\frac{14}{21}$ away from 0, and $\frac{9}{23}$ is $\frac{14}{23}$ away from 0. Since $\frac{14}{19}$ is the largest, we know that it is the farthest away from 0, and $\frac{14}{23}$ is the smallest, so it is the closest to 0. Therefore, our answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$ .

~monkey_land

Video Solution

https://youtu.be/pU1zjw--K8M ~savannahsolver

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

Page

Toolbox

Search