Difference between revisions of "2020 AMC 12A Problems/Problem 1"
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==Solution 1== | ==Solution 1== | ||
| − | If Carlos took 70% of the pie, (100 - 70) = 30% must be remaining. After Maria takes 1/ | + | If Carlos took <math>70\%</math> of the pie, <math>(100 - 70) = 30\%</math> must be remaining. After Maria takes <math>\frac{1}{3}</math> of the remaining <math>30\%, </math> |
| − | + | <math>1 - \frac{1}{3} = \frac{2}{3}</math> is left. | |
Therefore: | Therefore: | ||
| − | + | <math>\frac{3}{10} \cdot \frac{2}{3} = \frac{2}{10}= \boxed{\textbf{C) }20\%}</math> | |
| − | + | -Contributed by Awesome2.1, latex by quacker88 | |
| − | |||
| − | -Contributed by Awesome2.1 | ||
==See Also== | ==See Also== | ||
Revision as of 11:00, 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 of the pie, 
must be remaining. After Maria takes 
of the remaining 
is left.
Therefore:
-Contributed by Awesome2.1, latex by quacker88
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. 
