Difference between revisions of "2020 AMC 12A Problems/Problem 1"
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<math>\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 30\%\qquad\textbf{(E)}\ 35\%</math> | <math>\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 30\%\qquad\textbf{(E)}\ 35\%</math> | ||
| − | ==Solution== | + | ==Solution 1== |
If Carlos took 70% of the pie, (100 - 70) = 30% must be remaining. After Maria takes 1/3 of the remaining 30%, | If Carlos took 70% of the pie, (100 - 70) = 30% must be remaining. After Maria takes 1/3 of the remaining 30%, | ||
| Line 12: | Line 12: | ||
Therefore: | Therefore: | ||
| − | (3 / 10) * (2 / 3) = (2 / 10) = 20%, which is answer choice | + | (3 / 10) * (2 / 3) = (2 / 10) = 20%, which is answer choice C |
If anyone could add the latex to the numbers / expressions that would be really helpful! | If anyone could add the latex to the numbers / expressions that would be really helpful! | ||
Revision as of 09:34, 1 February 2020
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 70% of the pie, (100 - 70) = 30% must be remaining. After Maria takes 1/3 of the remaining 30%, (1 - 1/3) = 2/3 is left.
Therefore:
(3 / 10) * (2 / 3) = (2 / 10) = 20%, which is answer choice C
If anyone could add the latex to the numbers / expressions that would be really helpful!
-Contributed by Awesome2.1
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. 
