Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AMC 8 Problems/Problem 15"
(bdot)
(Tag: Undo)
Line 1: Line 1:
−
bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot
+
==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>
 +
 
 +
<math>\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300</math>
 +
 
 +
==Solution 1==
 +
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==
 +
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>.
 +
 
 +
==Solution 3==
 +
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>.
 +
 
 +
==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>.
 +
 
 +
==Video Solution==
 +
https://youtu.be/mjS-PHTw-GE
 +
 
 +
==See also==
 +
{{AMC8 box|year=2020|num-b=14|num-a=16}}
 +
{{MAA Notice}}

Revision as of 14:19, 23 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

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

Video Solution

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

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_15&oldid=138237"