Difference between revisions of "2020 AMC 8 Problems/Problem 15"
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==Video Solution== | ==Video Solution== | ||
https://youtu.be/xjwDsaRE_Wo | https://youtu.be/xjwDsaRE_Wo | ||
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+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/YnwkBZTv5Fw?t=665 | ||
+ | |||
+ | ~Interstigation | ||
==See also== | ==See also== | ||
{{AMC8 box|year=2020|num-b=14|num-a=16}} | {{AMC8 box|year=2020|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:33, 18 April 2021
Contents
Problem
Suppose of equals of What percentage of is
Solution 1
Since , multiplying the given condition by shows that is percent of .
Solution 2
Letting (without loss of generality), the condition becomes . Clearly, it follows that is of , so the answer is .
Solution 3
We have and , so . Solving for , we multiply by to give , so the answer is .
Solution 4
We are given , so we may assume without loss of generality that and . This means , and thus the answer is .
Video Solution
~savannahsolver
Video Solution
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=665
~Interstigation
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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.