Difference between revisions of "2019 AMC 8 Problems/Problem 3"
(→Solution 3) |
|||
Line 24: | Line 24: | ||
This is also similar to Problem 20 on the AMC 2012. | This is also similar to Problem 20 on the AMC 2012. | ||
+ | |||
+ | ==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>. | ||
==See Also== | ==See Also== |
Revision as of 01:21, 7 April 2020
Problem 3
Which of the following is the correct order of the fractions and
from least to greatest?
Solution 1
Consider subtracting 1 from each of the fractions. Our new fractions would then be and
. Since
, it follows that the answer is
-will3145
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 3
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.