Art of Problem Solving
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

2019 AMC 8 Problems/Problem 3: Difference between revisions

← Older editNewer edit →
Nivaar (talk | contribs)
editor
37 edits
Nivaar (talk | contribs)
editor
37 edits
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>.
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. That is, if  
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  
<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>.
<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>.


Revision as of 10:26, 23 January 2024

Problem 3

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)

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 fraction's 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, in order 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 back 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


Video Solution

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(Problem 1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

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. Error creating thumbnail: File missing

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.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_3&oldid=212394"