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| + | ==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 15:19, 23 November 2020
Problem
Suppose
of
equals
of
What percentage of
is
Solution 1
Since
, multiplying the given condition by
shows that
is
percent of
.
Solution 2
Letting
(without loss of generality), the condition becomes
. Clearly, it follows that
is
of
, so the answer is
.
Solution 3
We have
and
, so
. Solving for
, we multiply by
to give
, so the answer is
.
Solution 4
We are given
, so we may assume without loss of generality that
and
. This means
, and thus answer is
.
Video Solution
https://youtu.be/mjS-PHTw-GE
See also
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.