Difference between revisions of "2012 AMC 8 Problems/Problem 4"
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| + | ==Problem== | ||
Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. What fraction of the pizza did Peter eat? | Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. What fraction of the pizza did Peter eat? | ||
| − | <math> \textbf{(A)}\hspace{.05in}\frac{1}{24}\qquad\textbf{(B)}\hspace{.05in}\frac{1}{12}\qquad\textbf{(C)}\hspace{.05in}\frac{1}8\qquad\textbf{(D)}\hspace{.05in}\frac{1}6\qquad\textbf{(E)}\hspace{.05in}\frac{1}4 </math> | + | <math> \textbf{(A)}\hspace{.05in}\frac{1}{24}\qquad\textbf{(B)}\hspace{.05in}\frac{1}{12}\qquad\textbf{(C)}\hspace{.05in}\frac{1}{8}\qquad\textbf{(D)}\hspace{.05in}\frac{1}{6}\qquad\textbf{(E)}\hspace{.05in}\frac{1}{4} </math> |
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| + | ==Solution== | ||
| + | Peter ate <math>1 + \frac{1}{2} = \frac{3}{2}</math> slices. The pizza has <math> 12 </math> slices total. Taking the ratio of the amount of slices Peter ate to the amount of slices in the pizza, we find that Peter ate <math>\dfrac{\frac{3}{2}\text{ slices}}{12\text{ slices}} = \boxed{\textbf{(C)}\ \frac{1}{8}}</math> of the pizza. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2012|num-b=3|num-a=5}} | {{AMC8 box|year=2012|num-b=3|num-a=5}} | ||
| + | {{MAA Notice}} | ||
Revision as of 11:23, 23 November 2013
Problem
Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. What fraction of the pizza did Peter eat?
Solution
Peter ate 
slices. The pizza has 
slices total. Taking the ratio of the amount of slices Peter ate to the amount of slices in the pizza, we find that Peter ate 
of the pizza.
See Also
| 2012 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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. 
