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

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

m (→‎Solution 4(probably won't use this solution))
(→‎Problem 3)
Line 3: Line 3:
  
 
<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>
 
<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==
 
==Solution 1==
 
Each one is in the form <math>\frac{x+4}{x}</math> so we are really comparing <math>\frac{4}{11}, \frac{4}{15},</math> and <math>\frac{4}{13}</math> where you can see <math>\frac{4}{11}>\frac{4}{13}>\frac{4}{15}</math> so the answer is  <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>.
 
Each one is in the form <math>\frac{x+4}{x}</math> so we are really comparing <math>\frac{4}{11}, \frac{4}{15},</math> and <math>\frac{4}{13}</math> where you can see <math>\frac{4}{11}>\frac{4}{13}>\frac{4}{15}</math> so the answer is  <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>.

Revision as of 12:18, 2 November 2020

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

Each one is in the form $\frac{x+4}{x}$ so we are really comparing $\frac{4}{11}, \frac{4}{15},$ and $\frac{4}{13}$ where you can see $\frac{4}{11}>\frac{4}{13}>\frac{4}{15}$ so the answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$.

Solution 2

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 $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$.

-xMidnightFirex

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

Solution 3

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 AMC 2012.

Solution 4 (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.

Solution 5

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 $\boxed{\textbf{(E)}}$.

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

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

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=136417"