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

When $\frac{x}{y}<1$ and $z>0$ , $\frac{x+z}{y+z}>\frac{x}{y}$ . Hence, the answer is ${\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$ . ~ ryjs

This is also similar to Problem 3 on the AMC 8 2019, but with the rule switched.

Solution 4

By dividing, we see that 5/19 ≈ 0.26, 7/21 ≈ 0.33, and 9/23 ≈ 0.39. When we put this in order, $0.26$ < $0.33$ < $0.39$ . So our answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$ ~ math_genius_11

Solution 5

$\frac{5}{19}$ is very close to $\frac{1}{4}$ , so you can round it to that. Similarly, $\frac{7}{21} = \frac{1}{3}$ and $\frac{9}{23}$ can be rounded to $\frac{1}{2}$ , so our ordering is 1/4, 1/3, and 1/2, or $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$ .

Solution 6

The numbers are in form $\frac{x}{x+14}$ . Using quotient rule on $\frac{d}{dx}(\frac{x}{x+14})$ gives $\frac{14}{(x+14)^2}$ and this is positive. Because the derivative is always positive and the values of $x$ given by this question $(5, 7, 9)$ can be put on an interval that does not contain the critical point $x=-14$ , a greater $x$ implies a greater $\frac{x}{x+14}$ , thus giving us the answer of $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$ . ~lopkiloinm

Solution 7

We calculating each fraction subtracted from 1. Doing so, we find that $1 - \frac{5}{19} = \frac{14}{19}$ , $1 - \frac{7}{21} = \frac{14}{21}$ (we leave this unsimplified), and $1 - \frac{9}{23} = \frac{14}{23}$ . Clearly, $\frac{14}{19} > \frac {14}{21} > \frac{14}{23}$ . Using the fact that if $1 - a > 1 - b > 1 - c$ , then $a < b < c$ , we find that the correct ordering is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$ .

~ cxsmi

Video Solution

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

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=TpsuRedYOiM&t=250s

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

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