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Difference between revisions of "2019 AMC 8 Problems/Problem 3"

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== Solution 7 ==
 
== Solution 7 ==
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As you can see that the difference between the numerator and denominator is 4, 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>.
+
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>.
  
 
Note-this answer only works in some circumstances, and overall solution 6 is much better.
 
Note-this answer only works in some circumstances, and overall solution 6 is much better.

Latest revision as of 04:10, 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.

See also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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