Difference between revisions of "2019 AMC 12A Problems/Problem 1"
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The area of the larger pizza is <math>16\pi</math>, while the area of the smaller pizza is <math>9\pi</math>. Therefore, the larger pizza is <math>\frac{7\pi}{9\pi} \cdot 100\%</math> bigger than the smaller pizza. <math>\frac{7\pi}{9\pi} \cdot 100\% = 77.777....</math>, which is closest to <math>\boxed{\textbf{(E) }78}</math>. | The area of the larger pizza is <math>16\pi</math>, while the area of the smaller pizza is <math>9\pi</math>. Therefore, the larger pizza is <math>\frac{7\pi}{9\pi} \cdot 100\%</math> bigger than the smaller pizza. <math>\frac{7\pi}{9\pi} \cdot 100\% = 77.777....</math>, which is closest to <math>\boxed{\textbf{(E) }78}</math>. | ||
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==See Also== | ==See Also== | ||
Revision as of 21:46, 17 February 2019
Problem
The area of a pizza with radius 
is 
percent larger than the area of a pizza with radius 
inches. What is the integer closest to ?
Solution
The area of the larger pizza is 
, while the area of the smaller pizza is 
. Therefore, the larger pizza is 
bigger than the smaller pizza. 
, which is closest to 
.
See Also
| 2019 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. 
