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 answer is <math>\boxed{\textbf{(C) }75}</math>.
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WLOG, we can assume that <math>x=20</math> and <math>y=15</math> to make them equal. This means <math>\frac{y}{x}=\frac{15}{20}=\frac{75}{100}</math>, and thus answer is <math>\boxed{\textbf{(C) }75}</math>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 14:22, 25 November 2020

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

WLOG, we can assume that $x=20$ and $y=15$ to make them equal. This means $\frac{y}{x}=\frac{15}{20}=\frac{75}{100}$, and thus answer is $\boxed{\textbf{(C) }75}$.

Video Solution

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

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AJHSME/AMC 8 Problems and Solutions

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