Difference between revisions of "2019 AMC 8 Problems/Problem 3"
(→Solution 1) |
(→Solution 3) |
||
Line 22: | Line 22: | ||
This is also similar to Problem 20 on the AMC 2012. | This is also similar to Problem 20 on the AMC 2012. | ||
− | ==Solution | + | ==Solution 4== |
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>. | 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>. | ||
Revision as of 20:51, 24 June 2020
Problem 3
Which of the following is the correct order of the fractions and from least to greatest?
Solution 1
each one is 1+4/x so we are really comparing 4/11,4/15, and 4/13 where you can see 4/11>4/13>4/15 so the answer is E
Solution 2
We take a common denominator:
Since it follows that the answer is .
-xMidnightFirex
~ dolphin7 - I took your idea and made it an explanation.
Solution 2
When and , . Hence, the answer is . ~ ryjs
This is also similar to Problem 20 on the AMC 2012.
Solution 4
We use our insane mental calculator to find out that , , and . Thus, our answer is .
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.
Wot blitz