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

(Problem)
m (Solution 7)
 
(34 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,
+
Change <math>7/21</math> into <math>1/3</math>;
 +
<cmath>\frac{1}{3}\cdot\frac{5}{5}=\frac{5}{15}</cmath>
 +
<cmath>\frac{5}{15}>\frac{5}{19}</cmath>
 +
<cmath>\frac{7}{21}>\frac{5}{19}</cmath>
 +
And
 +
<cmath>\frac{1}{3}\cdot\frac{9}{9}=\frac{9}{27}</cmath>
 +
<cmath>\frac{9}{27}<\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>.
  
<math>1-\dfrac{5}{19}=\dfrac{14}{19}</math>
+
==Solution 3==
 +
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>.
 +
~ ryjs
  
<math>1-\dfrac{7}{21}=\dfrac{14}{21}</math>
+
This is also similar to Problem 3 on the AMC 8 2019, but with the rule switched.
  
<math>1-\dfrac{9}{23}=\dfrac{14}{23}</math>
+
==Solution 4==
  
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>.
+
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 5==
 +
 
 +
<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 6==
 +
 
 +
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 positive. Because the derivative is always positive and the values of <math>x</math> given by this question <math>(5, 7, 9)</math> can be put on an interval that does not contain the critical point <math>x=-14</math>, 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
 +
 
 +
==Solution 7==
 +
We calculating each fraction subtracted from 1. Doing so, we find that <math>1 - \frac{5}{19} = \frac{14}{19}</math>, <math>1 - \frac{7}{21} = \frac{14}{21}</math> (we leave this unsimplified), and <math>1 - \frac{9}{23} = \frac{14}{23}</math>. Clearly, <math>\frac{14}{19} > \frac {14}{21} > \frac{14}{23}</math>. Using the fact that if <math>1 - a > 1 - b > 1 - c</math>, then <math>a < b < c</math>, we find that the correct ordering is <math>\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}</math>.  
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
 +
 
 +
==Video Solution==
 +
https://youtu.be/pU1zjw--K8M ~savannahsolver
 +
 
 +
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=TpsuRedYOiM&t=250s
  
 
==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 18:34, 20 January 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

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

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

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