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

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==Solution 3==
 
==Solution 3==
 
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 talk:Aops-g5-gethsemanea2|talk]]) 07:41, 18 November 2020 (EST)
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--[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]]
  
 
==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 08:41, 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

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
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

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