Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AMC 12A Problems/Problem 1"

m (→‎Solution 1: Just checked the contribution history. Awesome2.1 and quacker88 are the correct users to give credits with.)
m (→‎Solution 2)
Line 15: Line 15:
  
 
==Solution 2==
 
==Solution 2==
−
Like solution 1, it is clear that there is <math>30\%</math> of the pie remaining. Since Maria takes <math>\frac{1}{3}</math> of the remainder, she takes <math>\frac{1}{3} \cdot 30\% = 10\%</math> meaning that there is <math>30\% - 10\% = \boxed{\textbf{(C)}\ 20\%}</math> left.
+
Like solution 1, it is clear that there is <math>30\%</math> of the pie remaining. Since Maria takes <math>\frac{1}{3}</math> of the remainder, she takes <math>\frac{1}{3} \cdot 30\% = 10\%,</math> meaning that there is <math>30\% - 10\% = \boxed{\textbf{(C)}\ 20\%}</math> left.
  
 
~DBlack2021
 
~DBlack2021

Revision as of 12:41, 27 August 2021

Problem

Carlos took $70\%$ of a whole pie. Maria took one third of the remainder. What portion of the whole pie was left?

$\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 30\%\qquad\textbf{(E)}\ 35\%$

Solution 1

If Carlos took $70\%$ of the pie, there must be $(100 - 70)\% = 30\%$ left. After Maria takes $\frac{1}{3}$ of the remaining $30\%, \ 1 - \frac{1}{3} = \frac{2}{3}$ of the remaining $30\%$ is left.

Therefore, the answer is $30\% \cdot \frac{2}{3} = \boxed{\textbf{(C)}\ 20\%}.$

~Awesome2.1 (Solution)

~quacker88 ($\LaTeX$ Adjustments)

Solution 2

Like solution 1, it is clear that there is $30\%$ of the pie remaining. Since Maria takes $\frac{1}{3}$ of the remainder, she takes $\frac{1}{3} \cdot 30\% = 10\%,$ meaning that there is $30\% - 10\% = \boxed{\textbf{(C)}\ 20\%}$ left.

~DBlack2021

Solution 3 (One Sentence)

We have \[\left(100\%-70\%\right)\cdot\left(1-\frac13\right)=30\%\cdot\frac23=\boxed{\textbf{(C)}\ 20\%}\] of the whole pie left.

~MRENTHUSIASM

Video Solution

https://youtu.be/qJF3G7_IDgc

~IceMatrix

Video Solution

https://www.youtube.com/watch?v=1fkJ2Mm55Ls

~The Power of Logic

See Also

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
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 AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_1&oldid=161009"