Difference between revisions of "2020 AMC 8 Problems/Problem 15"
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==Solution 1== | ==Solution 1== | ||
− | Since <math>20\% = \frac{1}{5}</math>, | + | 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== | ==Solution 2== | ||
Line 15: | Line 15: | ||
==Solution 4== | ==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>. | 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== | ==Video Solution== |
Latest revision as of 16:34, 26 January 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6
- 8 Video Solution by NiuniuMaths (Easy to understand!)
- 9 Video Solution by Math-X (First understand the problem!!!)
- 10 Video Solution (🚀Very Fast🚀)
- 11 Video Solution
- 12 Video Solution
- 13 Video Solution by Interstigation
- 14 See also
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 the answer is
.
Solution 5
of
is
, and
of
is
. We put
and
into an equation, creating
because
equals
. Solving for
, dividing
to both sides, we get
, so the answer is
.
Solution 6
of
can be written as
, or
.
of
can similarly be written as
, or
. So now,
. Using cross-multiplication, we can simplify the equation as:
. Dividing both sides by
, we get:
.
is the same thing as
, so the answer is
.
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🚀)
~Education, the Study of Everything
Video Solution
~savannahsolver
Video Solution
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.