Difference between revisions of "2020 AMC 8 Problems/Problem 15"
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− | ==Problem | + | ==Problem== |
Suppose <math>15\%</math> of <math>x</math> equals <math>20\%</math> of <math>y.</math> What percentage of <math>x</math> is <math>y?</math> | Suppose <math>15\%</math> of <math>x</math> equals <math>20\%</math> of <math>y.</math> What percentage of <math>x</math> is <math>y?</math> | ||
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==Solution 1== | ==Solution 1== | ||
− | + | Since <math>20\% = \frac{1}{5}</math>, multiplying the given condition by <math>5</math> shows that <math>y</math> is <math>15 \cdot 5 = \boxed{\textbf{(C) }75}</math> percent of <math>x</math>. | |
− | ==See also== | + | ==Solution 2== |
+ | Letting <math>x=100</math> (without loss of generality), the condition becomes <math>0.15\cdot 100 = 0.2\cdot y \Rightarrow 15 = \frac{y}{5} \Rightarrow y=75</math>. Clearly, it follows that <math>y</math> is <math>75\%</math> of <math>x</math>, so the answer is <math>\boxed{\textbf{(C) }75}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | We have <math>15\%=\frac{3}{20}</math> and <math>20\%=\frac{1}{5}</math>, so <math>\frac{3}{20}x=\frac{1}{5}y</math>. Solving for <math>y</math>, we multiply by <math>5</math> to give <math>y = \frac{15}{20}x = \frac{3}{4}x</math>, so the answer is <math>\boxed{\textbf{(C) }75}</math>. | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/xjwDsaRE_Wo | ||
+ | |||
+ | ==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 18:48, 6 January 2021
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
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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