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Difference between revisions of "2020 AMC 8 Problems/Problem 15"
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==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==
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Multiply by <math>5</math> to get <math>0.75x=y</math>. The <math>0.75</math> here can be converted to <math>75\%</math>. Therefore, <math>\boxed{\textbf{C}}</math> is the answer.
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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>.
  
 
==Solution 2==
 
==Solution 2==
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Letting <math>x=100</math>, our equation becomes <math>0.15\cdot 100 = 0.2\cdot y \implies 15 = \frac{y}{5} \implies y=75</math>. Clearly, <math>y</math> is <math>75\%</math> of <math>x</math> and the answer is <math>\boxed{\textbf{C}}</math>.<br>
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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>.
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~ junaidmansuri
 
  
 
==Solution 3==
 
==Solution 3==
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Let us transform the first sentence to an equation. <math>15\%=\frac3{20}</math> and <math>20\%=\frac15.</math> So, <math>\frac3{20}x=\frac15y.</math> Therefore, <math>\frac1{20}x=\frac1{15}y</math> and <math>x=\frac43y,</math> hence <math>\boxed{\textbf{(C) }75}</math>. <br>  
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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>.
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--[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] ([[User talk:Aops-g5-gethsemanea2|talk]]) 07:41, 18 November 2020 (EST)
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==Solution 4==
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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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==Solution 5==
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<math>15\%</math> of <math>x</math> is <math>0.15x</math>, and <math>20\%</math> of <math>y</math> is <math>0.20y</math>. We put <math>0.15x</math> and <math>0.20y</math> into an equation, creating <math>0.15x = 0.20y</math> because <math>0.15x</math> equals <math>0.20y</math>. Solving for <math>y</math>, dividing <math>0.2</math> to both sides, we get <math>y = \frac{15}{20}x = \frac{3}{4}x</math>, so the answer is <math>\boxed{\textbf{(C) }75}</math>.
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==Solution 6==
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<math>15\%</math> of <math>x</math> can be written as <math>\frac{15}{100}x</math>, or <math>\frac{15x}{100}</math>. <math>20\%</math> of <math>y</math> can similarly be written as <math>\frac{20}{100}y</math>, or <math>\frac{20y}{100}</math>. So now, <math>\frac{15x}{100} = \frac{20y}{100}</math>. Using cross-multiplication, we can simplify the equation as: <math>1500x = 2000y</math>. Dividing both sides by <math>500</math>, we get: <math>3x = 4y</math>. <math>\frac{3}{4}</math> is the same thing as <math>75\%</math>, so the answer is <math>\boxed{\textbf{(C) }75}</math>.
 +
 
 +
==Video Solution by NiuniuMaths (Easy to understand!)==
 +
https://www.youtube.com/watch?v=bHNrBwwUCMI
 +
 
 +
~NiuniuMaths
 +
 
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/UnVo6jZ3Wnk?si=fRl03D9Q1KAdDtXz&t=2346
 +
 
 +
~Math-X
 +
 
 +
==Video Solution (🚀Very Fast🚀)==
 +
https://youtu.be/8LyGag4DOzo
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution==
 +
https://youtu.be/mjS-PHTw-GE
 +
 
 +
~savannahsolver
 +
 
 +
==Video Solution==
 +
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}}

Latest revision as of 15:34, 26 January 2024

Problem

Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$

$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$

Solution 1

Since $20\% = \frac{1}{5}$, multiplying the given condition by $5$ shows that $y$ is $15 \cdot 5 = \boxed{\textbf{(C) }75}$ percent of $x$.

Solution 2

Letting $x=100$ (without loss of generality), the condition becomes $0.15\cdot 100 = 0.2\cdot y \Rightarrow 15 = \frac{y}{5} \Rightarrow y=75$. Clearly, it follows that $y$ is $75\%$ of $x$, so the answer is $\boxed{\textbf{(C) }75}$.

Solution 3

We have $15\%=\frac{3}{20}$ and $20\%=\frac{1}{5}$, so $\frac{3}{20}x=\frac{1}{5}y$. Solving for $y$, we multiply by $5$ to give $y = \frac{15}{20}x = \frac{3}{4}x$, so the answer is $\boxed{\textbf{(C) }75}$.

Solution 4

We are given $0.15x = 0.20y$, so we may assume without loss of generality that $x=20$ and $y=15$. This means $\frac{y}{x}=\frac{15}{20}=\frac{75}{100}$, and thus the answer is $\boxed{\textbf{(C) }75}$.

Solution 5

$15\%$ of $x$ is $0.15x$, and $20\%$ of $y$ is $0.20y$. We put $0.15x$ and $0.20y$ into an equation, creating $0.15x = 0.20y$ because $0.15x$ equals $0.20y$. Solving for $y$, dividing $0.2$ to both sides, we get $y = \frac{15}{20}x = \frac{3}{4}x$, so the answer is $\boxed{\textbf{(C) }75}$.

Solution 6

$15\%$ of $x$ can be written as $\frac{15}{100}x$, or $\frac{15x}{100}$. $20\%$ of $y$ can similarly be written as $\frac{20}{100}y$, or $\frac{20y}{100}$. So now, $\frac{15x}{100} = \frac{20y}{100}$. Using cross-multiplication, we can simplify the equation as: $1500x = 2000y$. Dividing both sides by $500$, we get: $3x = 4y$. $\frac{3}{4}$ is the same thing as $75\%$, so the answer is $\boxed{\textbf{(C) }75}$.

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=bHNrBwwUCMI

~NiuniuMaths

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=fRl03D9Q1KAdDtXz&t=2346

~Math-X

Video Solution (🚀Very Fast🚀)

https://youtu.be/8LyGag4DOzo

~Education, the Study of Everything

Video Solution

https://youtu.be/mjS-PHTw-GE

~savannahsolver

Video Solution

https://youtu.be/xjwDsaRE_Wo

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
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All AJHSME/AMC 8 Problems and Solutions

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