Difference between revisions of "2019 AMC 8 Problems/Problem 3"
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<math>\textbf{(A) }\frac{15}{11}< \frac{17}{13}< \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}< \frac{19}{15}<\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}<\frac{19}{15}<\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}<\frac{15}{11}<\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}<\frac{17}{13}<\frac{15}{11}</math> | <math>\textbf{(A) }\frac{15}{11}< \frac{17}{13}< \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}< \frac{19}{15}<\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}<\frac{19}{15}<\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}<\frac{15}{11}<\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}<\frac{17}{13}<\frac{15}{11}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Consider subtracting 1 from each of the fractions. Our new fractions would then be <math>\frac{4}{11}, \frac{4}{15},</math> and <math>\frac{4}{13}</math>. Since <math>\frac{4}{15}<\frac{4}{13}<\frac{4}{11}</math>, it follows that the answer is <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math> | Consider subtracting 1 from each of the fractions. Our new fractions would then be <math>\frac{4}{11}, \frac{4}{15},</math> and <math>\frac{4}{13}</math>. Since <math>\frac{4}{15}<\frac{4}{13}<\frac{4}{11}</math>, it follows that the answer is <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math> | ||
-will3145 | -will3145 | ||
| + | ==Solution 2== | ||
| + | You could change everything to a common denominator, which eventually gives us an answer of <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>. | ||
| + | |||
| + | -xMidnightFirex | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2019|num-b=2|num-a=4}} | {{AMC8 box|year=2019|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:48, 21 November 2019
Contents
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
You could change everything to a common denominator, which eventually gives us an answer of .
-xMidnightFirex
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. 
