2019 AMC 8 Problems/Problem 3: Difference between revisions

Latest revision as of 03:10, 12 August 2025

Problem

Which of the following is the correct order of the fractions $\frac{15}{11},\frac{19}{15},$ and $\frac{17}{13},$ from least to greatest?

$\textbf{(A) }\frac{15}{11}< \frac{17}{13}< \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}< \frac{19}{15}<\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}<\frac{19}{15}<\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}<\frac{15}{11}<\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}<\frac{17}{13}<\frac{15}{11}$

Solution 1 (Bashing/Butterfly Method)

We take a common denominator: $\frac{15}{11},\frac{19}{15}, \frac{17}{13} = \frac{15\cdot 15 \cdot 13}{11\cdot 15 \cdot 13},\frac{19 \cdot 11 \cdot 13}{15\cdot 11 \cdot 13}, \frac{17 \cdot 11 \cdot 15}{13\cdot 11 \cdot 15} = \frac{2925}{2145},\frac{2717}{2145},\frac{2805}{2145}.$

Since $2717<2805<2925$ it follows that the answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$ .

Another approach to this problem is using the properties of one fraction being greater than another, also known as the butterfly method. That is, if $\frac{a}{b}>\frac{c}{d}$ , then it must be true that $a * d$ is greater than $b * c$ . Using this approach, we can check for at least two distinct pairs of fractions and find out the greater one of those two, logically giving us the expected answer of $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$ .

-xMidnightFirex

~ dolphin7 - I took your idea and made it an explanation.

- Clearness by doulai1

- Alternate Solution by Nivaar

Solution 2

When $\frac{x}{y}>1$ and $z>0$ , $\frac{x+z}{y+z}<\frac{x}{y}$ . Hence, the answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$ . ~ ryjs

This is also similar to Problem 20 on the 2012 AMC 8.

Solution 3 (probably won't use this solution)

We use our insane mental calculator to find out that $\frac{15}{11} \approx 1.36$ , $\frac{19}{15} \approx 1.27$ , and $\frac{17}{13} \approx 1.31$ . Thus, our answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$ .

~~ by an insane math guy. ~~ random text that is here to distract you.

Solution 4

Suppose each fraction is expressed with denominator $2145$ : $\frac{2925}{2145}, \frac{2717}{2145}, \frac{2805}{2145}$ . Clearly $2717<2805<2925$ so the answer is $\boxed{\textbf{(E)}}$ .

Note: Duplicate of Solution 1

Solution 5 -SweetMango77

We notice that each of these fractions' numerator $-$ denominator $=4$ . If we take each of the fractions, and subtract $1$ from each, we get $\frac{4}{11}$ , $\frac{4}{15}$ , and $\frac{4}{13}$ . These are easy to order because the numerators are the same, we get $\frac{4}{15}<\frac{4}{13}<\frac{4}{11}$ . Because it is a subtraction by a constant, to order them, we keep the inequality signs to get $\boxed{\textbf{(E)}\;\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$ .

Solution 6

Adding on to Solution 5, we can turn each of the fractions $\frac{15}{11}$ , $\frac{17}{13}$ , and $\frac{19}{15}$ into $1$ $\frac{4}{11}$ , $1$ $\frac{4}{13}$ , and $1$ $\frac{4}{15}$ , respectively. We now subtract $1$ from each to get $\frac{4}{11}$ , $\frac{4}{15}$ , and $\frac{4}{13}$ . Since their numerators are all 4, this is easy because we know that $\frac{1}{15}<\frac{1}{13}<\frac{1}{11}$ and therefore $\frac{4}{15}<\frac{4}{13}<\frac{4}{11}$ . Reverting them to their original fractions, we can now see that the answer is $\boxed{\textbf{(E)}\;\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$ .

~by ChipmunkT

Solution 7

As you can see that the difference between the numerator and denominator is 4, you can guess the relative ratios and you can scale down and make it $\frac{9}{5},\frac{7}{3},$ and $\frac{5}{1},$ . Using this, you can very clearly tell that $\frac{5}{1},$ is the greatest and therefore $\frac{15}{11}$ , is the greatest. You can use the same method to find the other 2 orders. This gives the answer of $\boxed{\textbf{(E)}\;\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$ .

Note-this answer only works in some circumstances, and overall solution 6 is much better.

~ParticlePhysics

Video Solution by Math-X (First fully understand the problem!!!)

https://youtu.be/IgpayYB48C4?si=-TpVe8QyZbbc6yKr&t=266

~Math-X

The Learning Royal: https://youtu.be/IiFFDDITE6Q

Video Solution 2

Solution detailing how to solve the problem: https://www.youtube.com/watch?v=q27qEcr7TbQ&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=4

Video Solution 3

https://youtu.be/xNrhMAbaEcI

~savannahsolver

Video Solution

https://youtu.be/zI2f4GpQPIo

~Education, the Study of Everything

==Video Solution by The Power of Logic(1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

Butterfly Method

The butterfly method is a method where you multiply the denominator of the second fraction and multiply it by the numerator from the first fraction.

@@ Line 1: / Line 1: @@
-==Problem 3==
+==Problem==
 Which of the following is the correct order of the fractions <math>\frac{15}{11},\frac{19}{15},</math> and <math>\frac{17}{13},</math> from least to greatest?
 <math>\textbf{(A) }\frac{15}{11}< \frac{17}{13}< \frac{19}{15}  \qquad\textbf{(B) }\frac{15}{11}< \frac{19}{15}<\frac{17}{13}    \qquad\textbf{(C) }\frac{17}{13}<\frac{19}{15}<\frac{15}{11}    \qquad\textbf{(D) } \frac{19}{15}<\frac{15}{11}<\frac{17}{13}   \qquad\textbf{(E) }   \frac{19}{15}<\frac{17}{13}<\frac{15}{11}</math>
-==Solution 1 (Bashing)==
+==Solution 1 (Bashing/Butterfly Method)==
 We take a common denominator:
 <cmath>\frac{15}{11},\frac{19}{15}, \frac{17}{13} = \frac{15\cdot 15 \cdot 13}{11\cdot 15 \cdot 13},\frac{19 \cdot 11 \cdot 13}{15\cdot 11 \cdot 13}, \frac{17 \cdot 11 \cdot 15}{13\cdot 11 \cdot 15} = \frac{2925}{2145},\frac{2717}{2145},\frac{2805}{2145}.</cmath>
@@ Line 10: / Line 10: @@
 Since <math>2717<2805<2925</math> it follows that the answer is <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>.
-Another approach to this problem is using the properties of one fraction being greater than another, also known as the butterfly method. That is, if
+Another approach to this problem is using the properties of one fraction being greater than another, also known as the <b>butterfly method</b>. That is, if
 <math>\frac{a}{b}>\frac{c}{d}</math>, then it must be true that <math>a * d</math> is greater than <math>b * c</math>. Using this approach, we can check for <b>at least</b> two distinct pairs of fractions and find out the greater one of those two, logically giving us the expected answer of <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>.
@@ Line 39: / Line 39: @@
 ==Solution 5 -SweetMango77==
-We notice that each of these fraction's numerator <math>-</math> denominator <math>=4</math>. If we take each of the fractions, and subtract <math>1</math> from each, we get <math>\frac{4}{11}</math>, <math>\frac{4}{15}</math>, and <math>\frac{4}{13}</math>. These are easy to order because the numerators are the same, we get <math>\frac{4}{15}<\frac{4}{13}<\frac{4}{11}</math>. Because it is a subtraction by a constant, in order to order them, we keep the inequality signs to get <math>\boxed{\textbf{(E)}\;\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>.
+We notice that each of these fractions' numerator <math>-</math> denominator <math>=4</math>. If we take each of the fractions, and subtract <math>1</math> from each, we get <math>\frac{4}{11}</math>, <math>\frac{4}{15}</math>, and <math>\frac{4}{13}</math>. These are easy to order because the numerators are the same, we get <math>\frac{4}{15}<\frac{4}{13}<\frac{4}{11}</math>. Because it is a subtraction by a constant, to order them, we keep the inequality signs to get <math>\boxed{\textbf{(E)}\;\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>.
 == Solution 6 ==
-Adding on to Solution 5, we can turn each of the fractions <math>\frac{15}{11}</math>, <math>\frac{17}{13}</math>, and <math>\frac{19}{15}</math> into <math>1</math><math>\frac{4}{11}</math>, <math>1</math><math>\frac{4}{13}</math>, and <math>1</math><math>\frac{4}{15}</math>, respectively. We now subtract <math>1</math> from each to get <math>\frac{4}{11}</math>, <math>\frac{4}{15}</math>, and <math>\frac{4}{13}</math>. Since their numerators are all 4, this is easy because we know that <math>\frac{1}{15}<\frac{1}{13}<\frac{1}{11}</math> and therefore <math>\frac{4}{15}<\frac{4}{13}<\frac{4}{11}</math>. Reverting them back to their original fractions, we can now see that the answer is <math>\boxed{\textbf{(E)}\;\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>.
+Adding on to Solution 5, we can turn each of the fractions <math>\frac{15}{11}</math>, <math>\frac{17}{13}</math>, and <math>\frac{19}{15}</math> into <math>1</math><math>\frac{4}{11}</math>, <math>1</math><math>\frac{4}{13}</math>, and <math>1</math><math>\frac{4}{15}</math>, respectively. We now subtract <math>1</math> from each to get <math>\frac{4}{11}</math>, <math>\frac{4}{15}</math>, and <math>\frac{4}{13}</math>. Since their numerators are all 4, this is easy because we know that <math>\frac{1}{15}<\frac{1}{13}<\frac{1}{11}</math> and therefore <math>\frac{4}{15}<\frac{4}{13}<\frac{4}{11}</math>. Reverting them to their original fractions, we can now see that the answer is <math>\boxed{\textbf{(E)}\;\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>.
 ~by ChipmunkT
+== Solution 7 ==
+As you can see that the difference between the numerator and denominator is 4, you can guess the relative ratios and you can scale down and make it <math>\frac{9}{5},\frac{7}{3},</math> and <math>\frac{5}{1},</math>. Using this, you can very clearly tell that <math>\frac{5}{1},</math> is the greatest and therefore <math>\frac{15}{11}</math>, is the greatest. You can use the same method to find the other 2 orders. This gives the answer of <math>\boxed{\textbf{(E)}\;\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>.
-== Video Solution ==
+Note-this answer only works in some circumstances, and overall solution 6 is much better.
+~ParticlePhysics
 ==Video Solution by Math-X (First fully understand the problem!!!)==
@@ Line 70: / Line 74: @@
 ~Education, the Study of Everything
-====Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)==
+====Video Solution by The Power of Logic(1 to 25 Full Solution)==
 https://youtu.be/Xm4ZGND9WoY
 ~Hayabusa1
+==Butterfly Method==
+The butterfly method is a method where you multiply the denominator of the second fraction and multiply it by the numerator from the first fraction.
 ==See also==
 {{AMC8 box|year=2019|num-b=2|num-a=4}}
-{{MAA Notice}} The butterfly method is a method when you multiply the denominator of the second fraction and multiply it by the numerator from the first fraction.
+{{MAA Notice}}
+[[Category:Introductory Algebra Problems]]

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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

Page

Toolbox

Search