Difference between revisions of "2020 AMC 8 Problems/Problem 15"
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==Solution 4== | ==Solution 4== | ||
We are given <math>0.15x = 0.20y</math>, so we may assume without loss of generality that <math>x=20</math> and <math>y=15</math>. This means <math>\frac{y}{x}=\frac{15}{20}=\frac{75}{100}</math>, and thus the answer is <math>\boxed{\textbf{(C) }75}</math>. | We are given <math>0.15x = 0.20y</math>, so we may assume without loss of generality that <math>x=20</math> and <math>y=15</math>. This means <math>\frac{y}{x}=\frac{15}{20}=\frac{75}{100}</math>, and thus the answer is <math>\boxed{\textbf{(C) }75}</math>. | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/mjS-PHTw-GE | ||
| + | |||
| + | ~savannahsolver | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 17:20, 26 February 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
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. 
