Difference between revisions of "2020 AMC 12A Problems/Problem 1"
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==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 04:15, 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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.