Difference between revisions of "2020 AMC 12A Problems/Problem 1"
MRENTHUSIASM (talk | contribs) (Added in Solution 3.) |
MRENTHUSIASM (talk | contribs) m (→Solution 3 (One-Line Version)) |
||
| Line 23: | Line 23: | ||
==Solution 3 (One-Line Version)== | ==Solution 3 (One-Line Version)== | ||
| − | We have <cmath>\left(100\%-70\%\right)\cdot\left(1-\frac13\right)=30\%\cdot\frac23=\boxed{\textbf{(C)}\ 20\%} | + | We have <cmath>\left(100\%-70\%\right)\cdot\left(1-\frac13\right)=30\%\cdot\frac23=\boxed{\textbf{(C)}\ 20\%}</cmath> of the whole pie left. |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
Revision as of 03:15, 17 April 2021
Contents
Problem
Carlos took 
of a whole pie. Maria took one third of the remainder. What portion of the whole pie was left?
Solution 1
If Carlos took of the pie, there must be 
left. After Maria takes 
of the remaining 
is left.
Therefore:
-Contributed by YOur dad, one dude
Solution 2
Like solution 1, it is clear that there is 
of the pie remaining. Since Maria takes of the remainder, she takes 
meaning that there is 
left 
.
~DBlack2021
Solution 3 (One-Line Version)
We have ![\[\left(100\%-70\%\right)\cdot\left(1-\frac13\right)=30\%\cdot\frac23=\boxed{\textbf{(C)}\ 20\%}\]](http://latex.artofproblemsolving.com/0/c/e/0ce11cb93c4b54ee4c2318644ba2ce6fb8c76f81.png)
of the whole pie left.
~MRENTHUSIASM
Video Solution
~IceMatrix
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
