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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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 +
==Solution 5==
 +
<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>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 13:05, 4 January 2022

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 $\boxed{\textbf{(C) }75}$.

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
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

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