Difference between revisions of "2019 AMC 8 Problems/Problem 3"
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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:
![\[\frac{15}{11},\frac{19}{15}, \frac{17}{13} = \frac{15\cdot 15 \cdot 13}{11\cdot 15 \cdot 13},\frac{19 \cdot 11 \cdot 13}{15\cdot 11 \cdot 13}, \frac{17 \cdot 11 \cdot 15}{13\cdot 11 \cdot 15} = \frac{2925}{2145},\frac{2717}{2145},\frac{2805}{2145}.\]](http://latex.artofproblemsolving.com/1/6/6/166b8126850e1a57d7eb59016ca8a9ed6c1d148c.png)
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. 
