Difference between revisions of "2019 AMC 8 Problems/Problem 3"
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<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 (Bashing)== | + | ==Solution 1 (Bashing/Butterfly Method)== |
We take a common denominator: | 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> | <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>. | 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 <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>. | ||
-xMidnightFirex | -xMidnightFirex | ||
Line 15: | Line 18: | ||
- Clearness by doulai1 | - Clearness by doulai1 | ||
+ | |||
+ | - Alternate Solution by Nivaar | ||
==Solution 2== | ==Solution 2== |
Latest revision as of 11:28, 23 January 2024
Contents
- 1 Problem 3
- 2 Solution 1 (Bashing/Butterfly Method)
- 3 Solution 2
- 4 Solution 3 (probably won't use this solution)
- 5 Solution 4
- 6 Solution 5 -SweetMango77
- 7 Solution 6
- 8 Video Solution
- 9 Video Solution by Math-X (First fully understand the problem!!!)
- 10 Video Solution 2
- 11 Video Solution 3
- 12 Video Solution
- 13 ==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
- 14 See also
Problem 3
Which of the following is the correct order of the fractions and
from least to greatest?
Solution 1 (Bashing/Butterfly Method)
We take a common denominator:
Since it follows that the answer is
.
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
, then it must be true that
is greater than
. 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
.
-xMidnightFirex
~ dolphin7 - I took your idea and made it an explanation.
- Clearness by doulai1
- Alternate Solution by Nivaar
Solution 2
When and
,
. Hence, the answer is
.
~ 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 ,
, and
. Thus, our answer is
.
~~ by an insane math guy. ~~ random text that is here to distract you.
Solution 4
Suppose each fraction is expressed with denominator :
. Clearly
so the answer is
.
- Note: Duplicate of Solution 1
Solution 5 -SweetMango77
We notice that each of these fraction's numerator denominator
. If we take each of the fractions, and subtract
from each, we get
,
, and
. These are easy to order because the numerators are the same, we get
. Because it is a subtraction by a constant, in order to order them, we keep the inequality signs to get
.
Solution 6
Adding on to Solution 5, we can turn each of the fractions ,
, and
into
,
, and
, respectively. We now subtract
from each to get
,
, and
. Since their numerators are all 4, this is easy because we know that
and therefore
. Reverting them back to their original fractions, we can now see that the answer is
.
~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
~savannahsolver
Video Solution
~Education, the Study of Everything
==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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.