Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3)
(→‎Solution 3 (quick and easy))
 
(40 intermediate revisions by 11 users not shown)
Line 18: Line 18:
  
 
==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 ==
 
 
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>
Line 39: Line 27:
 
<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>.
 +
 +
==Solution 3 (quick and easy)==
 +
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>.
 +
 +
~monkey_land
 +
edited by ~NXC
 +
 +
==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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_8_Problems/Problem_20&oldid=234673"