Difference between revisions of "2012 AMC 8 Problems/Problem 4"
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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)} | + | <math> \textbf{(A)}~\frac{1}{24}\qquad\textbf{(B)}~\frac{1}{12}\qquad\textbf{(C)}~\frac{1}{8}\qquad\textbf{(D)}~\frac{1}{6}\qquad\textbf{(E)}~\frac{1}{4} </math> |
==Solution 1== | ==Solution 1== | ||
| Line 8: | Line 8: | ||
==Solution 2== | ==Solution 2== | ||
| − | + | An alternative way of solving this problem is to add the slices separately. When Peter takes a full slice, he takes <math>\frac{1}{12}</math> of the pizza. When he shares the slice with paul, he splits a slice of pizza into two, the equation is <math>\frac{1}{2}</math> of <math>\frac{1}{12}</math>, which is <math>\frac{1}{24}</math>. Adding gives: <cmath>\frac{1}{12}+\frac{1}{24}=\boxed{\textbf{(C)}~\frac{1}{8}}</cmath> | |
| + | |||
| + | ~SmartGrowth ~edited by megaboy6679 | ||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 00:14, 29 January 2023
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 1
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.
Solution 2
An alternative way of solving this problem is to add the slices separately. When Peter takes a full slice, he takes 
of the pizza. When he shares the slice with paul, he splits a slice of pizza into two, the equation is 
of , which is 
. Adding gives: ![\[\frac{1}{12}+\frac{1}{24}=\boxed{\textbf{(C)}~\frac{1}{8}}\]](http://latex.artofproblemsolving.com/e/5/6/e56fdcab5d0e08fd3918be722a87efef7b0be255.png)
~SmartGrowth ~edited by megaboy6679
Video Solution
https://youtu.be/WDRPwoRbpKo ~savannahsolver
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. 
