Difference between revisions of "2020 AMC 8 Problems/Problem 15"

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RedFIREtruck sucks
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==Problem==
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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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<math>\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300</math>
  
 
==Solution 1==
 
==Solution 1==
Do you know that taco bells make you go poo poo
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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==

Revision as of 16:49, 19 December 2021

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

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 (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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