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

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

@@ Line 18: / Line 18: @@
 ==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 2=
 Change <math>7/21</math> into <math>1/3</math>;
 <cmath>\frac{1}{3}\cdot\frac{5}{5}=\frac{5}{15}</cmath>
@@ Line 39: / Line 28: @@
 Therefore,  our answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>.
-==Solution 3==
+==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 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, <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 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 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
 ==Video Solution==
 https://youtu.be/pU1zjw--K8M ~savannahsolver
-==Video Solution by SpreadTheMathLove==
-https://www.youtube.com/watch?v=TpsuRedYOiM&t=250s
 ==See Also==
 {{AMC8 box|year=2012|num-b=19|num-a=21}}
 {{MAA Notice}}

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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