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2012 AMC 8 Problems/Problem 20: Difference between revisions

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==Solution 2==
==Solution 2==
Instead of finding the LCD, we can subtract each fraction from <math>1</math> to get a common numerator. Thus,
<math>1-\dfrac{5}{19}=\dfrac{14}{19}</math>
<math>1-\dfrac{7}{21}=\dfrac{14}{21}</math>
<math>1-\dfrac{9}{23}=\dfrac{14}{23}</math>
All three fractions have common numerator <math>14</math>. Now it is obvious the order of the fractions. <math>\dfrac{14}{19}>\dfrac{14}{21}>\dfrac{14}{23}\implies\dfrac{5}{19}<\dfrac{7}{21}<\dfrac{9}{23}</math>. Therefore,  our answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>.
==Solution 3==
Change <math>7/21</math> into <math>1/3</math>;
Change <math>7/21</math> into <math>1/3</math>;
<cmath>\frac{1}{3}\cdot\frac{5}{5}=\frac{5}{15}</cmath>
<cmath>\frac{1}{3}\cdot\frac{5}{5}=\frac{5}{15}</cmath>
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Therefore,  our answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>.
Therefore,  our answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>.


==Solution 4==
==Solution 3 (quick and easy)==
When <math>\frac{x}{y}<1</math> and <math>z>0</math>, <math>\frac{x+z}{y+z}>\frac{x}{y}</math>. Hence, the answer is <math>{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}</math>.
We know that <math>\frac{5}{19}</math> is <math>\frac{14}{19}</math> away from <math>1</math>, <math>\frac{7}{21}</math> is <math>\frac{14}{21}</math> away from <math>1</math>, and <math>\frac{9}{23}</math> is <math>\frac{14}{23}</math> away from <math>1</math>. Since <math>\frac{14}{19}</math> is the largest, we know that it is the farthest away from 0, and <math>\frac{14}{23}</math> is the smallest, so it is the closest to 0. Therefore, our answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>.
~ 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, <math>0.26</math> < <math>0.33</math> < <math>0.39</math>. So our answer is <math>\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}</math>
~ math_genius_11
 
==Solution 6==
 
<math>\frac{5}{19}</math> is very close to <math>\frac{1}{4}</math>, so you can round it to that. Similarly, <math>\frac{7}{21} = \frac{1}{3}</math> and <math>\frac{9}{23}</math> can be rounded to <math>\frac{1}{2}</math>, so our ordering is 1/4, 1/3, and 1/2, or <math>\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}</math>.


==Solution 7==
~monkey_land
edited by ~NXC


The numbers are in form <math>\frac{x}{x+14}</math>. Using quotient rule on <math>\frac{d}{dx}(\frac{x}{x+14})</math> gives <math>\frac{14}{(x+14)^2}</math> and this is always positive. Because it is always positive, a greater <math>x</math> implies a greater <math>\frac{x}{x+14}</math>, thus giving us the answer of <math>\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}</math>. ~lopkiloinm
==Video Solution==
https://youtu.be/pU1zjw--K8M ~savannahsolver


==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}}

Latest revision as of 14:01, 17 November 2024

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 (quick and easy)

We know that $\frac{5}{19}$ is $\frac{14}{19}$ away from $1$, $\frac{7}{21}$ is $\frac{14}{21}$ away from $1$, and $\frac{9}{23}$ is $\frac{14}{23}$ away from $1$. 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 edited by ~NXC

Video Solution

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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