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

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==Solution 1==
 
==Solution 1==
 
We set up the following equation based on the given information: <cmath>\frac{15x}{100}=\frac{20y}{100}</cmath> Solving for <math>x</math> yields <cmath>\frac{3x}{20}=\frac{y}{5}</cmath> <cmath>20y=15x</cmath> <cmath>x=1.\overline{3}y ==> D</cmath>
 
We set up the following equation based on the given information: <cmath>\frac{15x}{100}=\frac{20y}{100}</cmath> Solving for <math>x</math> yields <cmath>\frac{3x}{20}=\frac{y}{5}</cmath> <cmath>20y=15x</cmath> <cmath>x=1.\overline{3}y ==> D</cmath>
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==See also==
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{{AMC8 box|year=2020|num-b=14|num-a=16}}
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{{MAA Notice}}

Revision as of 00:21, 18 November 2020

Problem 15

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

We set up the following equation based on the given information: \[\frac{15x}{100}=\frac{20y}{100}\] Solving for $x$ yields \[\frac{3x}{20}=\frac{y}{5}\] \[20y=15x\] \[x=1.\overline{3}y ==> D\]

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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