2012 AMC 8 Problems/Problem 20: Difference between revisions
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==Solution 4== | ==Solution 4== | ||
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 {\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} | 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>{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}</math>. | ||
~ ryjs | ~ ryjs | ||
Revision as of 02:18, 24 December 2019
Problem
What is the correct ordering of the three numbers
,
, and
, in increasing order?
Solution 1
The value of
is
. Now we give all the fractions a common denominator.
Ordering the fractions from least to greatest, we find that they are in the order listed. Therefore, our final answer is
.
Solution 2
Instead of finding the LCD, we can subtract each fraction from
to get a common numerator. Thus,
All three fractions have common numerator
. Now it is obvious the order of the fractions.
. Therefore, our answer is
.
Solution 3
Change
into
;
And
Therefore, our answer is
.
Solution 4
When
and
,
. Hence, the answer is
.
~ ryjs
This is also similar to Problem 3 on the AMC 8 2019, but with the rule switched.
See Also
| 2012 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
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