Difference between revisions of "2019 AMC 8 Problems/Problem 3"
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Adding on to Solution 5, we can turn each of the fractions <math>\frac{15}{11}</math>, <math>\frac{17}{13}</math>, and <math>\frac{19}{15}</math> into <math>1</math><math>\frac{4}{11}</math>, <math>1</math><math>\frac{4}{13}</math>, and <math>1</math><math>\frac{4}{15}</math>, respectively. We now subtract <math>1</math> from each to get <math>\frac{4}{11}</math>, <math>\frac{4}{15}</math>, and <math>\frac{4}{13}</math>. Since their numerators are all 4, this is easy because we know that <math>\frac{1}{15}<\frac{1}{13}<\frac{1}{11}</math> and therefore <math>\frac{4}{15}<\frac{4}{13}<\frac{4}{11}</math>. Reverting them back to their original fractions, we can now see that the answer is <math>\boxed{\textbf{(E)}\;\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>. | Adding on to Solution 5, we can turn each of the fractions <math>\frac{15}{11}</math>, <math>\frac{17}{13}</math>, and <math>\frac{19}{15}</math> into <math>1</math><math>\frac{4}{11}</math>, <math>1</math><math>\frac{4}{13}</math>, and <math>1</math><math>\frac{4}{15}</math>, respectively. We now subtract <math>1</math> from each to get <math>\frac{4}{11}</math>, <math>\frac{4}{15}</math>, and <math>\frac{4}{13}</math>. Since their numerators are all 4, this is easy because we know that <math>\frac{1}{15}<\frac{1}{13}<\frac{1}{11}</math> and therefore <math>\frac{4}{15}<\frac{4}{13}<\frac{4}{11}</math>. Reverting them back to their original fractions, we can now see that the answer is <math>\boxed{\textbf{(E)}\;\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>. | ||
− | ~ ChipmunkT | + | ~by ChipmunkT |
== Video Solution == | == Video Solution == |
Revision as of 17:42, 1 January 2022
Contents
Problem 3
Which of the following is the correct order of the fractions and
from least to greatest?
Solution 1 (Bashing)
We take a common denominator:
Since it follows that the answer is
.
-xMidnightFirex
~ dolphin7 - I took your idea and made it an explanation.
- Clearness by doulai1
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.
Solution 4
Suppose each fraction is expressed with denominator :
. Clearly
so the answer is
.
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
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
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.