Difference between revisions of "2020 AMC 12A Problems/Problem 1"
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==Solution== | ==Solution== | ||
| − | If Carlos took 70% of the pie, 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%, |
| + | (1 - 1/3) = 2/3 is left. | ||
Therefore: | Therefore: | ||
| − | (3 / 10) * (2 / 3) = (2 / 10) = 20% | + | (3 / 10) * (2 / 3) = (2 / 10) = 20%, which is answer choice |
| + | |||
| + | If anyone could add the latex to the numbers / expressions that would be really helpful! | ||
| + | |||
| + | -Contributed by Awesome2.1 | ||
==See Also== | ==See Also== | ||
Revision as of 09:33, 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
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
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. 
