2020 AMC 8 Problems/Problem 15

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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