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"
(→‎Video Solution by Math-X (First understand the problem!!!))
 
(24 intermediate revisions by 15 users not shown)
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 the answer is <math>\boxed{\textbf{(C) }75}</math>.
 +
 
 +
==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>.
 +
 
 +
==Solution 6==
 +
<math>15\%</math> of <math>x</math> can be written as <math>\frac{15}{100}x</math>, or <math>\frac{15x}{100}</math>. <math>20\%</math> of <math>y</math> can similarly be written as <math>\frac{20}{100}y</math>, or <math>\frac{20y}{100}</math>. So now, <math>\frac{15x}{100} = \frac{20y}{100}</math>. Using cross-multiplication, we can simplify the equation as: <math>1500x = 2000y</math>. Dividing both sides by <math>500</math>, we get: <math>3x = 4y</math>. <math>\frac{3}{4}</math> is the same thing as <math>75\%</math>, so the answer is <math>\boxed{\textbf{(C) }75}</math>.
 +
 
 +
==Video Solution by NiuniuMaths (Easy to understand!)==
 +
https://www.youtube.com/watch?v=bHNrBwwUCMI
 +
 
 +
~NiuniuMaths
 +
 
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/UnVo6jZ3Wnk?si=fRl03D9Q1KAdDtXz&t=2346
 +
 
 +
~Math-X
 +
 
 +
==Video Solution (🚀Very Fast🚀)==
 +
https://youtu.be/8LyGag4DOzo
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==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==
 +
{{AMC8 box|year=2020|num-b=14|num-a=16}}
 +
{{MAA Notice}}

Latest revision as of 15:34, 26 January 2024

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

Solution 6

$15\%$ of $x$ can be written as $\frac{15}{100}x$, or $\frac{15x}{100}$. $20\%$ of $y$ can similarly be written as $\frac{20}{100}y$, or $\frac{20y}{100}$. So now, $\frac{15x}{100} = \frac{20y}{100}$. Using cross-multiplication, we can simplify the equation as: $1500x = 2000y$. Dividing both sides by $500$, we get: $3x = 4y$. $\frac{3}{4}$ is the same thing as $75\%$, so the answer is $\boxed{\textbf{(C) }75}$.

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=bHNrBwwUCMI

~NiuniuMaths

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=fRl03D9Q1KAdDtXz&t=2346

~Math-X

Video Solution (🚀Very Fast🚀)

https://youtu.be/8LyGag4DOzo

~Education, the Study of Everything

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

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=213189"