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

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<cmath>\frac{7}{21}<\frac{9}{23}</cmath>
 
<cmath>\frac{7}{21}<\frac{9}{23}</cmath>
 
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>.
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==Solution 3==
 
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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>.
 
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~ ryjs
 
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This is also similar to Problem 3 on the AMC 8 2019, but with the rule switched.
 
  
 
==Solution 4==
 
==Solution 4==

Revision as of 08:11, 16 July 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 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

Video Solution

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

Video Solution by SpreadTheMathLove

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

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

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