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Difference between revisions of "2020 AMC 8 Problems/Problem 15"
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Let us transform the first sentence to an equation. <math>15\%=\frac3{20}</math> and <math>20\%=\frac15.</math> So, <math>\frac3{20}x=\frac15y.</math> Therefore, <math>\frac1{20}x=\frac1{15}y</math> and <math>x=\frac43y,</math> hence <math>\boxed{\textbf{(C) }75}</math>. <br>  
 
Let us transform the first sentence to an equation. <math>15\%=\frac3{20}</math> and <math>20\%=\frac15.</math> So, <math>\frac3{20}x=\frac15y.</math> Therefore, <math>\frac1{20}x=\frac1{15}y</math> and <math>x=\frac43y,</math> hence <math>\boxed{\textbf{(C) }75}</math>. <br>  
 
--[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]]
 
--[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]]
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 +
==Solution 4==
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 +
We are given that <math>0.15x=0.20y</math>. Multiplying both sides by <math>100</math> and dividing by <math>20</math> tells us that <math>y = \frac 34x =0.75x=\textbf{(C) }75</math>.
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-franzliszt
  
 
==See also==
 
==See also==
 
{{AMC8 box|year=2020|num-b=14|num-a=16}}
 
{{AMC8 box|year=2020|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:59, 18 November 2020

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

Multiply by $5$ to get $0.75x=y$. The $0.75$ here can be converted to $75\%$. Therefore, $\boxed{\textbf{C}}$ is the answer.

Solution 2

Letting $x=100$, our equation becomes $0.15\cdot 100 = 0.2\cdot y \implies 15 = \frac{y}{5} \implies y=75$. Clearly, $y$ is $75\%$ of $x$ and the answer is $\boxed{\textbf{C}}$.
~ junaidmansuri

Solution 3

Let us transform the first sentence to an equation. $15\%=\frac3{20}$ and $20\%=\frac15.$ So, $\frac3{20}x=\frac15y.$ Therefore, $\frac1{20}x=\frac1{15}y$ and $x=\frac43y,$ hence $\boxed{\textbf{(C) }75}$.
--Aops-g5-gethsemanea2

Solution 4

We are given that $0.15x=0.20y$. Multiplying both sides by $100$ and dividing by $20$ tells us that $y = \frac 34x =0.75x=\textbf{(C) }75$.

-franzliszt

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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