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

(Problem 3)
 
(Solution 1)
 
(54 intermediate revisions by 30 users not shown)
Line 1: Line 1:
===Problem===
+
==Problem 3==
3. 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?  
+
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>
 
<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===
+
==Solution 1 (Bashing)==
Consider subtracting 1 from each of the fractions. Our new fractions would then be <math>\frac{4}{11}, \frac{4}{15}, and \frac{4}{13}</math>. Since <math>\frac{4}{15}<\frac{4}{13}<\frac{4}{11}</math>, it follows that the answer is <math>\boxed{\qquad\textbf{(E) }\frac{19}{15}<\frac{17}{13}<\frac{15}{11}</math>
+
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>
 +
 
 +
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>.
 +
 
 +
-xMidnightFirex
 +
 
 +
~ dolphin7 - I took your idea and made it an explanation.
 +
 
 +
- Clearness by doulai1
 +
 
 +
==Solution 2==
 +
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>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>.
 +
~ 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 <math>\frac{15}{11} \approx 1.36</math>, <math>\frac{19}{15} \approx 1.27</math>, and <math>\frac{17}{13} \approx 1.31</math>. Thus, our answer is <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>.
 +
 
 +
~~ by an insane math guy.
 +
 
 +
==Solution 4==
 +
Suppose each fraction is expressed with denominator <math>2145</math>: <math>\frac{2925}{2145}, \frac{2717}{2145}, \frac{2805}{2145}</math>. Clearly <math>2717<2805<2925</math> so the answer is <math>\boxed{\textbf{(E)}}</math>.
 +
 
 +
==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}{19}</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>.
 +
 
 +
== Video Solution ==
 +
 
 +
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=q27qEcr7TbQ&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=4
 +
 
 +
==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.

Latest revision as of 21:02, 18 September 2021

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

-xMidnightFirex

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

- Clearness by doulai1

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.

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

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 $\frac{4}{19}$. 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 $\boxed{\textbf{(E)}\;\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$.

Video Solution

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

See also

2019 AMC 8 (ProblemsAnswer KeyResources)
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.

Invalid username
Login to AoPS