Difference between revisions of "2019 AMC 8 Problems/Problem 3"
(→Solution 3) |
(→Solution 3) |
||
Line 22: | Line 22: | ||
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>. | 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 | ~ ryjs | ||
+ | |||
+ | This is also similar to Problem 20 on the AMC 2012. | ||
==See Also== | ==See Also== |
Revision as of 03:15, 24 December 2019
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.
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.