Difference between revisions of "2019 AMC 8 Problems/Problem 3"
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each one is 1+4/x so we are really comparing 4/11,4/15, and 4/13 where you can see 4/11>4/13>4/15 so the answer is E | each one is 1+4/x so we are really comparing 4/11,4/15, and 4/13 where you can see 4/11>4/13>4/15 so the answer is E | ||
− | ==Solution | + | ==Solution 2== |
We take a common denominator: | We take a common denominator: | ||
<cmath>\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}.</cmath> | <cmath>\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}.</cmath> |
Revision as of 21:51, 24 June 2020
Problem 3
Which of the following is the correct order of the fractions and
from least to greatest?
Solution 1
each one is 1+4/x so we are really comparing 4/11,4/15, and 4/13 where you can see 4/11>4/13>4/15 so the answer is E
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 2
When and
,
. Hence, the answer is
.
~ ryjs
This is also similar to Problem 20 on the AMC 2012.
Solution 3
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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