2012 AMC 8 Problems/Problem 20


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



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,




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

Solution 3

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 4

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 5

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 6

$\frac{5}{19}$ 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 7

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

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
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

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