Difference between revisions of "2020 AMC 8 Problems/Problem 15"
Cellsecret (talk | contribs) (→Video Solution) |
|||
Line 23: | Line 23: | ||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/xjwDsaRE_Wo | https://youtu.be/xjwDsaRE_Wo | ||
+ | |||
+ | ==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}} |
Revision as of 13:33, 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.