Difference between revisions of "2010 AMC 8 Problems/Problem 10"
m (→Problem 10) |
m (→Problem 10) |
||
| Line 1: | Line 1: | ||
| − | ==Problem | + | ==Problem== |
Six pepperoni circles will exactly fit across the diameter of a <math>12</math>-inch pizza when placed. If a total of <math>24</math> circles of pepperoni are placed on this pizza without overlap, what fraction of the pizza is covered by pepperoni? | Six pepperoni circles will exactly fit across the diameter of a <math>12</math>-inch pizza when placed. If a total of <math>24</math> circles of pepperoni are placed on this pizza without overlap, what fraction of the pizza is covered by pepperoni? | ||
Revision as of 20:14, 12 October 2018
Problem
Six pepperoni circles will exactly fit across the diameter of a 
-inch pizza when placed. If a total of 
circles of pepperoni are placed on this pizza without overlap, what fraction of the pizza is covered by pepperoni?
Solution
The pepperoni circles' diameter is 2, since 
. From that we see that the area of the circles of pepperoni is 
. The large pizza's area is 
. Therefore, the ratio is 
See Also
| 2010 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
